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CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 12 - MCQExams.com
CBSE
Class 12 Medical Physics
Alternating Current
Quiz 12
A sinusoidally varying potential difference has amplitude $$170 \,V$$.
What is its rms value?
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$$170 \,V$$
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$$170 \,V$$
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$$0$$
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$$-120 \,V$$
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$$-170 \,V$$
Explanation
$$\Delta V_{rms}=\Delta V_{max}/\sqrt{2}=170/\sqrt{2}=120 \,V$$
The reactance of a capacitor of capacitance C is X. If both the frequency and capacitance be doubled, then new reactance will be
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$$X$$
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$$2X$$
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$$4X$$
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$$\frac{X}{4}$$
Explanation
$$X_{c}=X$$
When both capacitance frequency are doubled the reactance will become
$$X_{C}^{'}=\frac{1}{2\pi \times (2f) \times (2C)}=\frac{1}{4} \times \frac{1}{2\pi fC}=\frac{X}{4}$$
In an A.C circuit, the resistance $$ R = 0.2\, \Omega$$. At a certain instant, $$V_A - V_B = 0.5\, V$$, $$ I = 0.5\, A$$, and the current is increasing at the rate of $$\dfrac{\Delta I}{\Delta t} = 8\,A/s$$. The inductance of the coil is:
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$$0.05\,H$$
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$$0.1\,H$$
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$$0.2\,H$$
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none of these
Explanation
Given $$I=0.5\ A$$
From Ohm's Law,
$$V_R=(0.5)\times (.2)$$
$$=0.1\ V$$
From Kirchoff's Voltage Law,
$$V_L=V_a-V_b-V_R = 0.5-0.1$$
$$=0.4\ V$$
For an inductor,
$$V_L=L\dfrac{di}{dt}$$
$$0.4=L\times8$$
$$L=0.05H$$
An AC source of angular frequency $$\omega $$ is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to $$\omega /3$$ (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency $$\omega $$ is
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$$\sqrt{\dfrac{3}{5}}$$
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$$\sqrt{\dfrac{5}{3}}$$
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$$\dfrac{3}{5}$$
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$$\dfrac{5}{3}$$
Explanation
$$I=\dfrac{V}{\sqrt{R^2+(\dfrac{1}{wc})^2}}$$, for w
for $$\dfrac{w}{3}, \dfrac{I}{2}=\dfrac{V}{\sqrt{R^2+(\dfrac{3}{wc})^2}}$$$$\dfrac{2}{\sqrt{R^2+\dfrac{9}{w^2C^2}}}=\dfrac{1}{\sqrt{R^2+\frac{1}{w^2C^2}}}$$$$4R^2+\dfrac{4}{w^2c^2}=R^2+\dfrac{9}{w^2c^2}$$
$$3R^2=\dfrac{5}{w^2c^2}$$
$$3R^2=5(x_c)^2$$ [at w]
$$\sqrt{\dfrac{3}{5}}=\dfrac{x_c}{R}$$
When a series combination of $$L$$ and $$R$$ are connected with a $$10V$$, $$50HZ$$ A.C. source, a current 1A flows in the circuit. The voltage leads the current by a phase angle of $$\pi/3$$ radian. Then the resistance is
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5$$\Omega $$
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10$$\Omega $$
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15$$\Omega $$
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20$$\Omega $$
Explanation
$$Vrms=10V, Irms=1A$$
i.e. $$tan\dfrac{\pi}{3}=\dfrac{x_L}{x_R}$$
$$\sqrt{3}=\dfrac{x_L}{R}$$
also $$\dfrac{10}{z}=1$$
$$10^2=R^2\times X_L^2$$
$$10^2=R^2+3^2$$
$$100=4R^2$$
$$R=5\Omega$$
A 100 volt A.C. source of frequency 500 hertz is connected to a L-C-R circuit with L $$=$$ 8.1 millihenry, C $$=$$ 12.5 microfarad and R $$=$$ 10 ohm, all connected in series. The potential difference across the resistance will be:
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10V
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100V
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50V
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500V
Explanation
Impedance of series RLC circuit is given by:
$$Z = \sqrt {R^2 + \left( 2 \pi f L - \dfrac{1}{2 \pi f C} \right)^2}$$
$$Z = \sqrt {10^2 + \left(2\pi \times 500 \times 8.1 \times10^{-3} - \dfrac{1}{2 \pi \times 500 \times 12.5 \times 10^{-6} } \right)^2}$$
$$Z \approx 10\ \Omega$$
Current flowing through the circuit is:
$$I = \dfrac{V}{Z} = 10A$$
Potential difference across the resistor, by ohm's Law is:
$$V_R = IR = 10 \times 10 = 100\ V$$
An AC source of variable frequency is applied across a series L-C-R circuit. At a frequency double the resonance frequency, the impedance is $$\sqrt{10}$$ times the minimum impedance. The inductive reactance is
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$$R$$
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$$2R$$
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$$3R$$
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$$4R$$
Explanation
Let $$\omega_0$$ be the resonance frequency
minimum impedance is at $$\omega_0$$
$$Z_{min}=X_R=R$$
At $$\omega = 2\omega_0$$,
$$Z=\sqrt{10}Z_{min}$$
$$\sqrt{{R}^2+(\omega L-\dfrac{1}{\omega C})^2}=\sqrt{10}\times R$$
$${R}^2+(2\omega_0 L-\dfrac{1}{2\omega_0 \times C})^2=10\times R^2$$
because at $$\omega_0 $$ , $$\omega_0 \times L=\dfrac{1}{\omega_0 C}$$ ,
$${R}^2+(2\omega_0 L-\dfrac{\omega_0 L}{2})^2=10\times R^2$$
$$(\dfrac{3}{2}\omega_ 0 L)^2=9\times R^2$$
$$\dfrac{3}{2}\times \omega_ 0 L=3\times R $$
$$X_L = \omega L = 2\omega_0 L=4 R$$ (at $$\omega=2\omega _0$$)
At a frequency $$\omega_{0} $$ the reactance of a certain capacitor equals that of a certain inductor. If frequency is changed to 2 $$\omega_{0} $$ . The ratio of reactance of the inductor to that of the capacitor is
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4 : 1
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$$\sqrt{2}:1$$
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$$1:2\sqrt{2}$$
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1 : 2
Explanation
at $$w_o, \dfrac{w_oL}{\dfrac{1}{w_oC}}=1$$
i.e. $$w_o^2LC=1$$
now at $$2w_o, \dfrac{2w_oL}{\dfrac{1}{2w_oC}}=\dfrac{X_L^1}{X_C^1}$$
$$4w_o^2LC=\dfrac{X_L^1}{X_C^1}$$
$$\dfrac{4}{1}=\dfrac{X_L^1}{X_C^1}$$
An ideal inductor takes a current of $$10 A$$ when connected to a $$125$$ $$V$$, $$50$$ $$Hz$$ AC supply. A pure resistor across the same source takes $$12.5$$ $$A$$. If the two are connected in series across a $$100\sqrt{2}V$$, $$ 40$$ $$Hz$$ supply, the current through the circuit will be
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$$10 A$$
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$$12.5 A$$
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$$20 A$$
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$$25 A$$
Explanation
Inductor
$$X_L=\dfrac{125}{10}=12.5\Omega$$
Resistor
$$X_R=\dfrac{125}{12.5}=10\Omega$$
$${ X }_{ L }'=12.5(\dfrac { { \omega }_{ 2 } }{ { \omega }_{ 1 } } )\Omega =12.5(\dfrac { 40 }{ 50 } )\Omega =10\quad \Omega$$
both are connected in series
$$z=\sqrt{10^2+(10)^2}$$
$$=10\sqrt { 2 } \Omega $$
$$Vrms=100\sqrt{2} \ V$$
$$Irms=\dfrac { 100\sqrt { 2 } }{ 10\sqrt { 2 } } =10A$$
A circuit containing resistance $$R_{1}$$, Inductance $$L_{1}$$ and capacitance $$C_{1}$$ connected in series resonates at the same frequency $$n$$ as a second combination of $$R_{2}$$, $$L_{2}$$ and$$C_{2}$$. If the two are connected in series, then the circuit will resonate at:
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$$n$$
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$$2n$$
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$$\sqrt{\dfrac{L_{2}C_{2}}{L_{1}C_{1}}}$$
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$$\sqrt{\dfrac{L_{1}C_{1}}{L_{2}C_{2}}}$$
Explanation
Condition of resonance $$\omega = \dfrac {1} { \sqrt {LC}}$$
So $$\omega_{1} = \dfrac {1} { \sqrt {L_{1}C_{1}}}$$
$$\omega_{2} = \dfrac {1} { \sqrt {L_{2}C_{2}}}$$
Since both frequency are same $$L_{1}C_{1} = L_{2}C_{2}$$
Equivalent capacitance of $$C_{1} \ and \ C_{2}$$ in series = $$ \dfrac {C_{1} C_{2}} { C_{1}+c_{2}}$$
Equivalent inductance of $$L_{1}$$ and $$L_{2} = L_{1} + L_{2}$$
Resonance for combination = $$ \sqrt{\dfrac { (C_{1} +C_{2})} { C_{1}C_{2}(L_{1}+L_{2})}}$$
Using $$L_{1}C_{1} = L_{2}C_{2}$$, resonance of combination = $$\dfrac{1}{\sqrt{L_{1}C_{1}}}$$
So the frequency of resonance remains same = $$n$$
In the given circuit, $$R$$ is a pure resistor, $$L$$ is a pure inductor, S is a 100V, 50Hz AC source and A is an AC ammeter. With either $$K_1$$ or $$K_2$$ alone closed, the ammeter reading is $$I$$. If the source is changed to 100 V, 100 Hz, the ammeter reading with $$K_1$$ alone closed and with $$K_2$$ alone closed will be respectively
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$$I , \dfrac{I}{ 2}$$
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$$I , 2I$$
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$$2I , I$$
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$$2I , \dfrac{I }{ 2}$$
Explanation
$$I=\dfrac{100}{R}=\dfrac{100}{2\pi(50)(L)}$$
now at $$100 Hz$$
new I when $$K_1$$ is close only
$$I_1=\dfrac{100}{R}=I$$
and new I when only $$K_2$$ is closed
$$I_2=\dfrac{100}{2\pi\times (100)\times L}=\dfrac{I}{2}$$
The self-inductance of the motor of an electric fan is 10H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of : (Take $${\pi}^2$$ = 10)
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$$4\mu F$$
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$$8\mu F$$
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$$1 \mu F$$
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$$2\mu F$$
Explanation
$$L=10H, f=50 Hz, w=50\times 2\pi$$
$$w=\dfrac{1}{\sqrt{LC}}$$
$$C=\dfrac{1}{Lw^2}$$
$$C=\dfrac{1}{10\times \pi^2\times 10^4}$$
$$=10^{-6}f$$
$$=1\mu F$$
A 120V, 60Hz a.c. power is connected 800$$\Omega $$ non-inductive resistance and unknown capacitance in series. The voltage drop across the resistance is found to be 102V, then voltage drop across capacitor is
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8V
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102V
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63V
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55V
Explanation
For RC circuits,
$$V^2 = V_R^2 +V_C^2$$
$$120^2 = 102^2 + V_C^2$$
$$V_C^2 = 63.214 \approx\ 63 V$$
An alternating e.m.f. of $$E= 100 \sin(100 \pi$$t ) is connected to a choke of negligible resistance and current oscillations of amplitude 1 A are produced. The inductance of the choke should be
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$$100 H$$
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$$\dfrac{1}{\pi } \ H$$
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$$1 H$$
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$$\pi H$$
Explanation
Angular frequency $$\omega=100\pi$$
Reactance $$X_L=100\pi\times L$$
$$I_o=1 \ A$$
$$E_o=100 \ V$$
$$\therefore \dfrac{E_o}{X_L}= \dfrac{100}{100\pi\times L}=1$$
$$L=\dfrac{1}{\pi}H$$
A coil has an inductance of 0.7H and is joined in series with a resistance of 220 $$\Omega $$ . When an alternating e.m.f. of 220V at 50 c.p.s. is applied to it, then the wattless component of the current in the circuit is
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5 ampere
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0.5 ampere
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0.7 ampere
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7 ampere
Explanation
Impedance of the circuit is given by:
$$Z = \sqrt {R^2 + (2 \pi f L)^2}$$
$$Z = \sqrt {220^2 + (2 \pi \times 50 \times 0.7)^2}$$
$$Z \approx 220\sqrt 2\ \Omega$$
Power factor is given by:
$$\cos \phi = \dfrac{R}{Z} = \dfrac{\pi}{4}$$
Total current in the circuit flowing is:
$$I = \dfrac{V}{Z} = \dfrac{220}{220\sqrt 2} = \dfrac{1}{\sqrt 2} A$$
Wattless component of current in the circuit is:
$$I_a = I \sin (\phi)$$
$$I_a = \dfrac{1}{\sqrt 2} \times \dfrac{1}{\sqrt 2} = 0.5 A$$
In the given circuit the readings of the voltmeter $$V_1$$ and the ammeter $$A$$ are:
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220 V, 2.2 A
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110 V; 1.1A
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220 V, 1.1 A
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110 V; 2.2 A
Explanation
$$w=100\pi\Omega$$
we observe that $$w^2=\dfrac{1}{LC}$$
i.e. this frequency is resonance frequency
all the voltage drop across resistor.
$$\therefore v_1=220V$$
and $$i=\dfrac{V_1}{R}$$
$$=\dfrac{220}{100}$$
$$=2.2A$$
Assertion: A resistance is connected to an ac source. Now a capacitor is included in the series circuit. The average power absorbed by the resistance will remain same.
Reason: By including a capacitor or an inductor
in the circuit average power across resistor does
not change.
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A and R both are true and R is correct explanation of A
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A and R both are true but R is not the correct explanation of A
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A is true R is false
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A is false and R is true
Explanation
After connecting the capacitor Irms will charge because impedance is changed.
$$\therefore$$ A is false
A certain choke coil of negligible resistance draws a current of 8A, when connected to a supply of 100 volts, at 50Hz. A certain non-inductive resistance, under the same conditions carries a current of 10 A. If the two are transferred to a supply system working at 150 V, at 40 Hz, the total current they will take if joined in series is
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1.06 A
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10.6 A
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0.106 A
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100.6 A
Explanation
In choke coil, $$I=8A, Erms=100V$$
$$\omega=100\pi \ rad/s$$
$$\dfrac{100}{100\pi\times L}=8=L=\dfrac{1}{8\pi}H$$
In resistance, $$I=10 \ amp$$
$$\therefore R=\dfrac{100}{10}=10 \ \Omega$$
New source
V=150 V, f= 40 Hz,
$$\omega'=40\times 2\pi$$
Connected in series, $$z=\sqrt{10^2+(\dfrac{80\pi}{8\pi})}$$
$$=10\sqrt{2}\Omega$$
$$Irms=\dfrac{150}{10\sqrt{2}}$$
$$=10.61 \ A$$
A capacitor has a resistance of $$1200 M\Omega$$ and capacitance of $$22 \mu F$$. When connected to an AC supply of frequency 80Hz, the alternating voltage supply required to drive a current of 10 virtual amperes is
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$$904\sqrt{2} V$$
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904 V
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$$904/\sqrt{2} V$$
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452 V
Explanation
$$X_c=1200\times 10^6\Omega$$
$$C=22\mu F$$
$$w=160\pi\ rad/s$$
$$\dfrac{V_o}{X_c}=10A$$
$$V_o=\dfrac{10}{160\pi(22\times 10^{-6})}=904.29V$$
The potential difference across a $$2 H$$ inductor as a function of time is shown in figure. At time $$t=0$$, current is zero. Current at $$t=2$$ second is:
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$$1 A$$
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$$3 A$$
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$$4 A$$
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$$5 A$$
Explanation
$$t=0, i=0$$
$$t=2, i=?$$
$$L=2H$$
$$V=L\dfrac{dI}{dt}$$
$$V=2\dfrac{dI}{dt}$$
$$\int_{0}^{2}Vdt=2\int_{0}^{I}dI$$
area of shaded portion
$$\dfrac{1}{2}\times 2\times 10=2\times I$$
$$I=5A$$
A sinusoidal voltage $$e = 150\ sin (100$$ $$\pi t) V$$ is applied to a series connection of resistance $$R$$ and inductance $$L$$.
Select TRUE statement/s from the following :
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Steady state current leads the voltage by $$30^{\circ}$$
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The instantaneous voltages across the resistance and inductance become equal at $$t = $$ $$\dfrac{1}{300}s$$
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Voltage leads the current by $$30^{\circ}$$
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The maximum current in the circuit is $$0.75\ A$$
An inductor of inductance $$L= 200\ \text{mH}$$ and the resistors $$R_{1}$$ and $$R_{2}$$, each of which is equal to $$4\ \Omega $$, are connected to a battery of emf $$12\ \text{V}$$ through a switch $$S$$ as shown in the Figure. The switch $$S$$ is closed and after the steady state is reached, the switch $$S$$ is opened. The current in $$R_{1}$$ then is $$($$in $$A)$$
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$$6e^{-20t}\ \text{from B to A}$$
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$$3e^{-40t}\ \text{from A to B}$$
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$$3e^{-40t}\ \text{from B to A}$$
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$$6e^{-20t}\ \text{from A to B}$$
Explanation
When S is closed at t = 0 s, current in $$LR_{2}$$ branch
$$i_{2}=3\left ( 1-e^{-\dfrac{R_{2}}{L}t} \right ),\dfrac{R_{2}}{L}=\dfrac{4}{2\times 10^{-1}}=20\ s^{-1}$$
$$=3(1-e^{-20t})A$$
Current i, through $$R_{1}=3A$$
When S is opened, current in R$$_{1}$$ reduces to zero just after opening.
But in $$L-R_{2}$$, it decreases exponentially
$$\therefore$$ Current through $$R_{1}=3e^{-40t}$$ A, from B to A
The frequency at which the impedance of the circuit becomes maximum is
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$$\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}+\dfrac{R^{2}}{L^{2}}}$$
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$$\dfrac{1}{2\pi }\dfrac{1}{\sqrt{LC}}$$
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$$\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}}$$
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$$\dfrac{1}{2\pi }\dfrac{R}{L}$$
Explanation
Magnitude of admittance $$Y=\dfrac{1}{Z}$$ is given by
$$|\displaystyle
\mathrm{Y}|=\dfrac{[\mathrm{R}^{2}\mathrm{X}_{\mathrm{c}}^{2}+(\mathrm{R}^{2}+\mathrm{X}_{L}^{2}-X_{L}X_{\mathrm{c}})^{2}]}{\mathrm{X}_{\mathrm{c}}(\mathrm{R}^{2}+\mathrm{X}_{L}^{2})}\dfrac{1}{2}$$
Now, $$|Z|$$ is maximum or $$|Y|$$ is minimum at $$\omega =\omega _{0}$$ such that
$$R^{2}+X_{L}^{2}-X_{L}X_{C}=0$$
where $$X_L = w_oL$$ and $$X_C = \dfrac{1}{w_o C}$$
Or $$R^{2}+\omega _{0}^{2}L^{2}=\omega _{0}L\times \dfrac{1}{\omega _{0}C}=\dfrac{L}{C}$$
or $$\omega _{0}^{2}=\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}$$
or $$\omega _{0}=\sqrt{\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}}$$
Hence resonance frequency $$f_{0}=\dfrac{w_o}{2\pi}=\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}}$$
A series RLC circuit is made as shown in the figure with an AC source of 60 V, 20 Hz. Then
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the rms current through the resistor R is 4.2 A
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the effective potential difference between P and Q should be 42 V.
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the instantaneous current leads the source voltage by $$45^{\circ}$$
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the instantaneous current lags behind the applied voltage by $$45^{\circ}$$
Explanation
$$Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{100+100}=10\sqrt{2}\Omega $$
$$\therefore I=\dfrac{60}{10\sqrt{2}}=6\times 0.707=4.2A$$
$$\therefore V_{R}=4.2\times 10=42V $$
The phase difference $$\phi =tan^{-1}\left ( \dfrac{X_{L}-X_{C}}{R} \right )=tan^{-1}1=45^{\circ}$$
As $$X_L > X_C$$ , Source voltage leads the current by $$45^{\circ}$$
The reading shown in AC Voltmeter V when S is moved to B is:
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120 V
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100 V
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140 V
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160 V
Explanation
Phase difference $$=\phi =45$$
$$cos \phi = \dfrac{R}{\sqrt {X_L^2+R^2}} = \dfrac{1}{\sqrt 2}$$
Solving, $$ X_L=R$$
$$\implies X_L=100\Omega$$
Impedance is $$Z = \sqrt {R^2+X_L^2} = 100\sqrt 2$$
Current in the circuit:
$$I_o=\dfrac{e}{Z} = \dfrac{200}{(\sqrt{2})100}=\sqrt{2}A$$
$$I_{rms}=1A$$
AC voltmeter shows RMS reading. Hence,
$$V_{rms}=I_{rms} R =(100\Omega)(1A)$$
$$=100V$$
In the circuit diagram shown, $$X_{C}=100\Omega , X_{L}=200 \Omega$$ & $$R=100 \Omega .$$ The effective current through the source is:
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2 A
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$$2 \sqrt{2} A$$
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0.5 A
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$$\sqrt{0.4} A$$
Explanation
$$I_{R}=\dfrac{V}{R}=\dfrac{200}{100}=2A$$
$${I}'=\dfrac{V}{X_{L}-X_{C}}=\dfrac{200}{100}=2A$$
$$I=\sqrt{I{_{R}}^{2}+{I}'^{2}}=2\sqrt{2} Amp.$$
In a series L, C circuit , which of the following represents variation of magnitude of reactance (X) with frequency (f) ?
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0%
0%
0%
Explanation
$$X=\left ( \omega L- \dfrac{1}{\omega C} \right ).$$
This is of the form $$y=x-\dfrac{1}{x}$$ which gives us the graph as shown.
For the values of frequency between 0 and 1 we take its absolute value and take the mirror image of the part which is below the x axis.
A current source sends a current i $$= i_0 cos (\omega t).$$When connected across an unknown load gives a voltage output of, $$v = v_0 sin (\omega t + \pi/4)$$across that load. Then voltage across the current source may be brought in phase with the current through it by
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connecting an inductor in series with the load
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connecting a capacitor in series with the load
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connecting an inductor in parallel with the load
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connecting a capacitor in parallel with the load.
Explanation
Here, current leads the voltage by a phase of $$\pi/4$$.
Thus unknown load is a capacitor.
The current can be brought in phase with current through it by connecting it i series with an inductor of same impedance as that of the capacitor.
Switch S is closed at t =After sufficiently long time an iron rod is inserted into the inductor L. Then, the bulb :
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glows more brightly
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gets dimmer
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glows with the same brightness
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gets momentarily dimmer and then glows more brightly
Explanation
When the iron rod is inserted into the inductor, the inductance of the coil increases.
As a result potential difference across the inductor increases, potential difference across the resistor( bulb ) decreases, so the bulb becomes dimmer.
The instantaneous potential difference between points A and B is (Phase angle is $$37^0$$)
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$$8 \sin (50\pi t+37\dfrac{\pi }{180})$$
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$$8 \sin (50\pi t-37\dfrac{\pi }{180})$$
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$$10 \sin (50\pi t )$$
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$$10 \cos (50\pi t )$$
Explanation
Impedance $$z=\sqrt{(8-2)^{2}+(8)^{2}}= 10\Omega $$
current lags voltage by $$37^{\circ}$$,then
$$i=\dfrac{10}{10} \sin (50\pi t-37^{\circ} )$$
$$V_{AB}=i\times R=8\sin (50\pi t-37^{\circ} )$$
The current flowing through the resistor in a series LCR a.c. circuit, is $$I =\varepsilon / R. $$
Now the inductor and capacitor are connected in parallel and joined in series with the resistor as shown in figure. The current in the circuit is now. (Symbols have their usual meaning)
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equal to I
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more than I
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less than I
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zero
Explanation
In figure (1) $$ Z = R$$
$$ X_{L} = X_{C}$$
If $$ I_{1} = \dfrac{V_{0}}{X_{C}} sin(\omega t+\pi/2)$$
then $$ I_{2} = \dfrac{V_{0}}{X_{L}} sin(\omega t-\pi/2)$$
and Total current in circuit two $$ I = I_{1} + I_{2}=0$$
An ac voltage source $$V = V_0\sin(\omega t)$$ is connected across resistance $$R$$ and capacitance $$C$$ as shown in figure. It is given that $$R = \dfrac{1}{\omega C}$$. The peak current is $$I_0$$ . If the angular frequency of the voltage source is changed to $$\dfrac{\omega}{\sqrt{3}}$$ then the new peak current in the circuit is:
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$$\dfrac{I_0}{2}$$
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$$\dfrac{I_0}{\sqrt{2}}$$
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$$\dfrac{I_0}{\sqrt{3}}$$
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$$\dfrac{I_0}{3}$$
Explanation
The peak value of the current is:
$$I_0 = \dfrac{V_0}{\sqrt{R^2+\dfrac{1}{{\omega}^2C^2}}} = \dfrac{V_0}{\sqrt{2}R}$$
When the angular frequency is changed to $$\dfrac{\omega}{\sqrt{3}}$$, the new peak value is:
$$I'_0 = \dfrac{V_0}{\sqrt{R^2+\dfrac{3}{{\omega}^2C^2}}}=\dfrac{V_0}{\sqrt{4R^2}}=\dfrac{V_0}{2R}$$
$$\therefore I'_0 = \dfrac{I_0}{\sqrt{2}}$$
A coil has resistance $$30\ \text{ohm}$$ and inductive reactance $$20\ \text{Ohm at}\ 50\ \text{Hz}$$ frequency. If an ac source, of $$200\ \text{volt},\ 100\ \text{Hz},$$ is connected across the coil, the current in the coil will be :
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$$4.0\ \text{A}$$
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$$8.0\ \text{A}$$
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$$\dfrac {20}{\sqrt {13}}\ \text{A}$$
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$$2.0\ \text{A}$$
Explanation
For a given coil
R:Resistance$$=30\Omega $$
$${ X }_{ L }:$$ Inductive Reactance $$=\omega L\Omega =\left( 2\pi F \right) L\Omega $$
At $$50㎐,{ X }_{ L }=20\Omega $$
$$\therefore 2\pi \times 50\times L=20$$
$$\therefore L=\cfrac { 20 }{ 100\pi } $$
At $$100㎐$$
$${ X }_{ L }=2\pi \times 100\times L=2\pi \times 100\times \cfrac { 20 }{ 100\pi } =40\Omega $$
At $$100㎐$$
$$z=impedance=\sqrt { { R }^{ 2 }+{ { X }_{ L } }^{ 2 } } $$
$$\therefore Z=\sqrt { { 40 }^{ 2 }+{ 30 }^{ 2 } } =50\Omega $$
Current at $$200V,100㎐$$ is given as
$$I=\cfrac { V }{ z } =\cfrac { 200 }{ 50 } =4A$$
Therefore option A is correct
Regarding the given circuit, the correct statement is:
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($$V_a - V_b$$) is increasing with time
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($$V_a - V_b$$) is decreasing with time
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($$V_a - V_b) = $$10V
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($$V_a - V_b) =$$ zero
Explanation
$$ Q_{c} = 5(1-e^{-\dfrac{t}{2}})$$
$$ I_{L} = 5(1-e^{-\dfrac{t}{2}})$$
$$ V_{a} - V_{b} = 2I_{L} - \dfrac{Q _{c}}{0.5}$$
$$ = 10(1-e^{-\dfrac{t}{2}}) - \dfrac{5}{0.5}(1-e^{-\dfrac{t}{2}})$$
$$ = 0$$
Calculate $$\omega$$ and $$T$$.
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$$6.28\times{10}^{-3}rad/s$$; $${10}^{-3}s$$
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$$6.28\times{10}^{3}rad/s$$; $${10}^{-3}s$$
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$$7.28\times{10}^{3}rad/s$$; $${10}^{-3}s$$
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$$7.28\times{10}^{-3}rad/s$$; $${10}^{-3}s$$
Explanation
Oscillation frequency $$f = 10^3 \ Hz$$
Angular frequency $$w = 2\pi f = 2\pi\times 10^3 = 6.28\times 10^3 \ rad/s$$
Time period $$T = \dfrac{1}{f} = \dfrac{1}{10^3} = 10^{-3} \ s^{-1}$$
When $$100$$ volt $$DC$$ source is applied across a coil, a current of $$1A$$ flows through it. When $$100V$$ $$AC$$ source of $$50Hz$$ is applied to the same coil, only $$0.5A$$ current flows. Calculate the inductance of the coil.
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$$\displaystyle (\pi/ \sqrt 3)H$$
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$$\displaystyle (\sqrt 3/ \pi)H$$
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$$\displaystyle (2/\pi)H$$
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None of these
Explanation
$$\displaystyle I_{dc}=\frac{V_{dc}}{R}$$
$$\displaystyle I=\frac{100}{R}\Rightarrow R=100\Omega$$
$$\displaystyle I_{ac}=\frac{V_{ac}}{\sqrt{R^2+X^2_L}}$$
$$\displaystyle 0.5=\frac{100}{\sqrt{(100)^2+X^2_L}}$$
or $$\displaystyle X_L=100\sqrt 3 \Omega =(2\pi fL)$$
$$\displaystyle \therefore L=\frac{100\sqrt 3}{2\pi f}$$
$$\displaystyle =\frac{100\sqrt 3}{2\pi (50)}$$
$$\displaystyle = \left(\frac{\sqrt 3}{\pi}\right)H$$
A coil, a capacitor and an $$AC$$ source of rms voltage $$24V$$ are connected in series. By varying the frequency of the source, a maximum rms current of $$6A$$ is observed. If coil is connected to a dc battery of emf $$12$$ volt and internal resistance $$4\Omega$$, then current through it in steady state is :
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$$2.4A$$
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$$1.8A$$
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$$1.5A$$
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$$1.2A$$
Explanation
$$\displaystyle I_{rms}=\frac{V}{R}$$ (at resonance)
$$\displaystyle 6=\frac{24}{R}$$
$$\displaystyle \therefore R=4\Omega$$
$$\displaystyle I_{dc}=\frac{V}{R+r}=\frac{12}{4+4}=1.5A$$
Identify the graph which correctly represents the variation of capacitive reactance $$X_C$$ with frequency.
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0%
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Explanation
$$X_C=\displaystyle \frac{1}{\omega C}=\frac{1}{2\pi fC}$$
or $$X_C\propto \displaystyle\frac{1}{f}$$
i.e., $$X_C$$ versus $$f$$ graph is a rectangular hyperbola.
A $$50Hz$$ $$AC$$ source of $$20V$$ is connected across $$R$$ and $$C$$ as shown in figure. The voltage across $$R$$ is $$12V$$. The voltage across $$C$$ is
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$$8V$$
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$$16V$$
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$$10V$$
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Not possible to determine unless value of $$R$$ and $$C$$ are given.
Explanation
$$\displaystyle V_C=\sqrt{V^2-V^2_R}=\sqrt{(20)^2-(12)^2}$$
$$=16V$$
The frequency of oscillation of current in the inductance is
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$$\dfrac {1}{3\sqrt {LC}}$$
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$$\dfrac {1}{6\pi \sqrt {LC}}$$
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$$\dfrac {1}{\sqrt {LC}}$$
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$$\dfrac {1}{3\pi \sqrt {LC}}$$
Explanation
Step 1: Find equivalent inductance
Here, both inductor are connected in series
So, Equivalent inductance is given by
$$\mathrm{L}_{\mathrm{eq}}=\mathrm{L}+2\mathrm{~ L}=3L$$
$$\textbf{Step 2: Find equivalent Capacitance}$$
Here, both capacitors are connected in parallel
So, Equivalent capacitance is given by
$$\mathrm{C}_{\mathrm{eq}}=\mathrm{C}+2\mathrm{C}=3C$$
$$\textbf{Step 3: Find the frequency of oscillation.}$$
Hence, Frequency of oscillation is
$$\Rightarrow\mathrm{f}=\frac{1}{2\pi\sqrt{\mathrm{L}_{\mathrm{eq}}\mathrm{C}_{\mathrm{eq}}}}$$
$$\Rightarrow\frac{1}{6\pi\sqrt{\mathrm{LC}}}$$
Therefore, The correct option is 'B' is
$$\frac{1}{6 \pi \sqrt{L C}}$$.
Consider an $$L-C-R$$ circuit as shown in figure, with an $$AC$$ source of peak value $$V_0$$ and angular frequency $$\omega$$. Then the peak value of current through the $$AC$$ source is :
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$$\displaystyle \frac{V_0}{\sqrt{\displaystyle \frac{1}{R^2}+\left(\omega L-\displaystyle\frac{1}{\omega C}\right)^2}}$$
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$$\displaystyle V_0 \sqrt {1/ {R^2}+\left(\omega C-\displaystyle \frac{1}{\omega L}\right)^2 }$$
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$$\displaystyle \frac{V_0}{\sqrt {R^2+\left(\omega L-\displaystyle \frac{1}{\omega C}\right)}^2}$$
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None of these.
Explanation
As R, L , C are in parallel, equivalent impedance is given by:
$$ \dfrac{1}{Z} = \dfrac{1}{Z_R } + \dfrac{1}{Z_L} + \dfrac{1}{Z_C} = \dfrac{1}{R} + \dfrac{1}{j L \omega} + j C \omega $$
$$ \left | \dfrac{1}{Z} \right| = \left | \dfrac{1}{R} + \dfrac{1}{j L \omega} + j C \omega \right | = \sqrt{ \dfrac{1}{R^2} + \left( \dfrac{1}{L \omega} - C \omega \right)^2 } $$
Peak value of current is given by:
$$ I_{peak} = \left | \dfrac{V_0}{Z} \right |= { V_0}{ \sqrt{ \dfrac{1}{R^2} + \left( \dfrac{1}{L \omega} - C \omega \right)^2 } }$$
When an alternating voltage of $$220V$$ is applied across a device $$P$$, a current of $$0.25A$$ flows through the circuit and it leads the applied voltage by an angle $$\pi/2$$ radian. When the same voltage source is connected across another device $$Q$$, the same current is observed in the circuit but in phase with the applied voltage. What is the current when the same source is connected across a series combination of $$P$$ and $$Q$$?
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$$\displaystyle \frac{1}{4\sqrt 2}A$$ lagging in phase by $$\pi/4$$ with voltage.
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$$\displaystyle \frac{1}{4\sqrt 2}A$$ leading in phase by $$\pi/4$$ with voltage.
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$$\displaystyle \frac{1}{\sqrt 2}A$$ leading in phase by $$\pi/4$$ with voltage.
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$$\displaystyle \frac{1}{4\sqrt 2}A$$ leading in phase by $$\pi/2$$ with voltage.
Explanation
In first case, $$\displaystyle X_C=\frac{V}{I}=\frac{220}{0.25}=880\Omega$$
In the second case, $$\displaystyle R=\frac{V}{I}=\frac{220}{0.25}=880\Omega$$
In the combination of $$P$$ and $$Q$$
$$\displaystyle \tan\phi=\frac{X_C}{R}=1$$
$$\therefore \phi=45^o$$
Since the circuit is capacitive, current leads the voltage. Further,
$$\displaystyle Z=\sqrt{R^2+X^2_C}=880\sqrt 2\Omega$$
$$\displaystyle I=\frac{V}{Z}=\frac{220}{880\sqrt 2}=\frac{1}{4\sqrt 2}A$$
A signal generator supplies a sine wave of $$200V$$, $$5kHz$$ to the circuit as shown in the figure. Then choose the wrong statement:
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The current in the resistive branch is $$0.2A$$.
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The current in the capacitive branch is $$0.126A$$.
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Total line current is $$=0.283A$$.
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Current in both the branches is same.
Explanation
$$\displaystyle I_R=\frac{V_{rms}}{R}=\frac{200}{100}=0.2A$$
$$\displaystyle X_C=\frac{1}{2\pi fC}=\frac{1}{(2\pi)(5\times 10^3)\left(\frac{1}{\pi}\times 10^{-6}\right)}$$
$$=100\Omega$$
$$\therefore \displaystyle I_C=\frac{V_{rms}}{X_C}=\frac{200}{100}=0.2A$$
$$I_C$$ is $$90^o$$ ahead of the applied voltage and $$I_R$$ is in phase with the applied voltage. Hence, there is a phase difference of $$90^o$$ between $$I_R$$ and $$I_C$$ too.
$$\therefore \displaystyle I=\sqrt{I^2_R+I^2_C}$$
$$=\displaystyle \sqrt{(0.2)^2+(0.2)^2}$$
$$=0.283A$$
An $$LCR$$ circuit contains resistance of $$100\Omega$$ and a supply of $$200\space V$$ at $$300\space rad/s$$ angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by $$60^{\small\circ}$$. If on the other hand, only inductor is taken out, the current leads by $$60^{\small\circ{}}$$ with the applied voltage. The current flowing in the circuit is
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$$1\space A$$
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$$1.5\space A$$
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$$2\space A$$
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$$2.5\space A$$
Explanation
if only capacitance is remove,
$$ Z=R+jL\omega \quad \Rightarrow \quad \left| Z \right| cos(60)=R=100$$
$$\Rightarrow \quad \left| Z \right| =200\quad \Omega$$
$$\Rightarrow \quad L\omega =\left| Z \right| sin(60)=173.205 $$
if only inductor is taken out,
$$ Z=R-j\frac { 1 }{ C\omega } \quad \Rightarrow \quad \left| Z \right| cos(-60)=R=100$$
$$\Rightarrow \quad \left| Z \right| =200\quad \Omega$$
$$\Rightarrow \quad \frac { 1 }{ C\omega } =\left| Z \right| sin(-60)=-173.205 $$
therefore actual impedance= $$ R+jL\omega - j \frac{1}{C\omega} = 100 \Omega $$
therefore current in circuit=$$\dfrac{200 V }{100 \Omega} = 2A $$
Resonance occurs in a series $$LCR$$ circuit when the frequency of the applied emf is $$1000\space Hz$$.
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When frequency $$= 900\space Hz$$, then the current through the voltage source will be ahead of emf of the source
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The impedance of the circuit is minimum at $$f = 1000\space Hz$$
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At only resonance the voltages across $$L$$ and $$C$$ differ in phase by $$180^{\small\circ}$$
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If the value of $$C$$ is doubled, resonance occurs at $$f = 2000\space Hz$$
Explanation
$$ f_0 = \frac{1}{2 \pi \sqrt{LC}} = 1000 Hz $$
when $$ f=900Hz< f_0 \Rightarrow \omega < \frac{1}{\sqrt{ LC}} \Rightarrow L \omega < \frac{ 1}{C \omega } \Rightarrow X_L < X_C \Rightarrow$$ circuit is capacitive. therefore the current leads the voltage of the source.
total impedance of circuit = $$ \sqrt { R^2 + (X_L - X_C )^2} \Rightarrow $$ impedance will be minimum when $$ X_L-X_C=0 $$ i.e., impedance will be minimum at resonance frequency of $$ 1000 Hz $$
voltage across L and C is always differ in phase by 180 degree .
resonant frequency when C is doubled = $$ \frac{1} { 2 \pi \sqrt{ L 2 C }} = \frac{1} {\sqrt{2}} \frac{1}{2 \pi \sqrt{LC}} = \frac{1000}{\sqrt{2} } Hz $$
A capacitor of $$10\space \mu F$$ and an inductor of $$1\space H$$ are joined in series. An ac of $$50\space Hz$$ is applied to this combination. What is the impedance of the combination?
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$$\displaystyle\frac{5(\pi^2 - 5)}{\pi}\space \Omega$$
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$$\displaystyle\frac{10(10 - \pi^2)}{\pi}\space \Omega$$
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$$\displaystyle\frac{10(\pi^2 - 5)}{\pi}\space \Omega$$
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$$\displaystyle\frac{5(10 - \pi^2)}{\pi}\space \Omega$$
Explanation
$$ X_L = L \omega = L 2 \pi f = 100 \pi \Omega $$
$$ X_C = \frac{ 1 } { C \omega } = \frac{ 1 } { C 2 \pi f } = 1000/ \pi \Omega $$
impedance =$$ \left | X_L - X_C \right | = 100 \dfrac{ 10- {\pi}^2 }{\pi} \Omega $$
A transmitter transmits at a wavelength of $$300\space m$$. A condenser of capacitance $$2.4\space \mu F$$ is being used. The value of the inductance for the resonant circuit is approximately
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$$10^{-4}\space H$$
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$$10^{-6}\space H$$
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$$10^{-8}\space H$$
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$$10^{-10}\space H$$
Explanation
$$ f = \dfrac{c}{ \lambda} = 3 \times 10^8 / 300 $$
$$ \omega = 2\pi f = \dfrac{1}{\sqrt{L C }} $$
$$ L = \dfrac{ 1}{C 4 {\pi}^{2} f^2 }= 10^{-8} H $$
Impedance of box $$P$$ at the above frequency is
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$$70\Omega$$
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$$77\Omega$$
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$$90\Omega$$
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$$100\Omega$$
Explanation
$$\displaystyle X_C=\frac{1}{\omega C}=\frac{1}{\left(\displaystyle \frac{10^5}{7}\right)(10^{-6})}=70\Omega$$
$$\displaystyle Z_P=\sqrt{R^2_P+X^2_C}$$
$$=\displaystyle \sqrt{(32)^2+(70)^2}=77\Omega$$
The angular frequency for which he detects maximum current in the circuit is
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$$10^5/7 rad/s$$
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$$10^4 rad/s$$
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$$10^5 rad/s$$
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$$10^4/7 rad/s$$
Explanation
$$\displaystyle \omega =\frac{1}{\sqrt {LC}}$$
$$=\displaystyle \frac{1}{\sqrt{4.9\times 10^{-3}\times 10^{-6}}}$$
$$\displaystyle =\frac{10^5}{7} rad/s$$
A choke coil of resistance $$5\Omega$$ and inductance $$0.6\space H$$ is in series with a capacitance of $$10\space \mu F$$. If a voltage of $$200\space V$$ is applied and the frequency is adjusted to resonance, the current and voltage across the inductance and capacitance are $$I_0, V_0$$ and $$V_1$$ respectively. We have
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$$I_0 = 40\space A$$
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$$V_0 = 9.8\space kV$$
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$$V_1 = 9.8\space kV$$
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$$V_1 = 19.6\space kV$$
Explanation
At resonance current, $$ I_0= \dfrac{ V}{R} = \dfrac{ 200} {5 } = 40 A $$
At resonance frequency voltage across L $$= V_0 $$, and voltage across C $$= V_1 $$ are always equal and out of phase by 180 degree.
Therefore, if $$f=65$$ Hz , $$ V_0 = j I_0 L \omega =I_0 L 2 \pi f = j 9800 V $$
$$ V_1 = -j \dfrac{I_0}{ C \omega } = -j \dfrac{ I_0}{ C 2 \pi f} = -j \ 9800 V $$
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Practice Class 12 Medical Physics Quiz Questions and Answers
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