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CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 8 - MCQExams.com
CBSE
Class 12 Medical Physics
Alternating Current
Quiz 8
A resistor R, inductor L and a capacitor C are connected in series to an oscillator of frequency v. If the resonant frequency is $$v_r$$, then the current lags behind the voltage, when:
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$$v = 0$$
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$$v < v_r$$
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$$v>v_r$$
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$$v=v_r$$
Explanation
The current will lag behind the voltage when reactance of inductor is more than the reactance of capacitor i.e.,
$$\omega L>\dfrac{1}{\omega c}$$
or $$\omega > \dfrac{1}{\sqrt{LC}}\ or \ 2\pi v>\dfrac{1}{\sqrt{LC}}$$
$$v>\dfrac{1}{2\pi \sqrt{LC}}\ or\ v>v_r$$
where $$v_r$$ is the resonant frequency.
In L-C-R circuit, $$f=\dfrac { 50 }{ \pi } Hz$$, $$V=50 V$$, $$R=300\Omega$$. If $$L=1H$$ and $$C=20\mu C$$, then the voltage across capacitor is :
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$$50 V$$
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$$20 V$$
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Zero
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$$30 V$$
Explanation
For an L-C-R circuit, the impedance $$\left( Z \right) $$ is given by
$$\because Z=\sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 } } $$
where, $${ X }_{ L }=\omega L=2\pi { f }_{ 1 }$$
and $${ X }_{ C }=\dfrac { 1 }{ \omega C } =\dfrac { 1 }{ 2\pi fC } $$
Given, $$f=\dfrac { 50 }{ \pi } Hz$$, $$R=300\Omega $$ and $$L=1H$$
and $$C=20\mu C=20\times { 10 }^{ -6 }C$$
$$Z=\sqrt { { \left( 300 \right) }^{ 2 }+\left( 2\pi \times \dfrac { 50 }{ \pi } \times 1-\dfrac { 1 }{ 2\pi \times \dfrac { 50 }{ \pi } \times 20\times { 10 }^{ -6 } } \right) } $$
$$Z=\sqrt { 90000+{ \left( 100-500 \right) }^{ 2 } } $$
$$\Rightarrow Z=\sqrt { 90000+160000 } =\sqrt { 250000 } $$
$$\Rightarrow Z=500\Omega $$
Hence, the current in the circuit is given by
$$i=\dfrac { V }{ Z } =\dfrac { 50 }{ 500 } =0.1A$$
Voltage across capacitor is
$${ V }_{ C }=i{ X }_{ C }=\dfrac { i }{ 2\pi fC } =\dfrac { 0.1 }{ 2\pi \times \dfrac { 50 }{ \pi } \times 20\times { 10 }^{ -4 } } $$
$$=\dfrac { 0.1\times { 10 }^{ 6 } }{ 100\times 20 } \Rightarrow { V }_{ C }=50V$$
A coil of inductive reactance $${ 1 }/{ \sqrt { 3 } }\Omega$$ and resistance $$ 1\Omega $$ is connected to $$200 V$$, $$50 Hz$$ A.C. supply. The time lag between maximum voltage and current is
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$$\dfrac { 1 }{ 600 } s$$
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$$\dfrac { 1 }{ 200 } s$$
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$$\dfrac { 1 }{ 300 } s$$
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$$\dfrac { 1 }{ 500 } s$$
Explanation
$$\omega=2\pi f=314 rad/s$$
$$tan\phi=\dfrac{1}{1/\sqrt{3}}\rightarrow \phi=\dfrac{\pi}{6}$$
$$\omega t=\dfrac{\pi}{6}$$
$$t=\dfrac{1}{600}s$$
An inductive coil has a resistance of 100 When an AC signal of frequency 1000 Hz is applied to the coil, the voltage leads the current by $$45^o$$. The inductance of the coil is
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$$\dfrac{1}{10 \pi}$$
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$$\dfrac{1}{20 \pi}$$
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$$\dfrac{1}{40 \pi}$$
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$$\dfrac{1}{60 \pi}$$
Explanation
Given : $$f = 1000$$ Hz
Resistance of inductive coil $$X_L = 100\Omega$$
Thus inductance of coil $$L = \dfrac{X_L}{2\pi f}$$
$$\therefore$$ $$L = \dfrac{100}{2\pi \times 1000} = \dfrac{1}{20\pi }$$ Henry
An AC source is connected in parallel with an L-C-R circuit as shown. Let $$\displaystyle { I }_{ S },{ I }_{ L },{ I }_{ C }$$ and $$\displaystyle { I }_{ R }$$ denote the currents through and $$\displaystyle { V }_{ S },{ V }_{ L },{ V }_{ C }$$ and $$\displaystyle { V }_{ R }$$ voltages across the corresponding components. Then :
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$$\displaystyle { I }_{ S }={ I }_{ L }+{ I }_{ C }+{ I }_{ R }$$
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$$\displaystyle { V }_{ S }={ V }_{ L }+{ V }_{ C }+{ V }_{ R }$$
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$$\displaystyle \left( { I }_{ L },{ I }_{ C },{ I }_{ R } \right) <{ I }_{ S }$$
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$$\displaystyle { I }_{ L },{ I }_{ C }$$ may be greater than $$\displaystyle { I }_{ S }$$
Explanation
For parallel RLC circuit,
$$I_S = \sqrt{I_R^2 + (I_L-I_C)^2}$$
In case of resonance, $$I_L = I_C$$
So, $$I_S = I_R$$
Here $$I_L$$ and $$I_C$$ can be much greater than $$I_R$$
Option D is correct.
At inductance $$1\ H$$ is connected in series with an $$AC$$ source of $$220\ V$$ and $$50\ Hz$$. The inductive resistance (in ohm) is :
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$$2\pi$$
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$$50\pi$$
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$$100\pi$$
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$$1000\pi$$
Explanation
Inductive reactants $$X_{L} = wL$$
$$= 2\pi vL$$
$$= 2\pi \times 50\times 1$$
$$= 100\pi$$.
In an LCR circuit having L$$=8$$H, C$$=0.5\mu F$$ and $$R=100\Omega$$ in series, the resonance frequency rad/s is?
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$$600$$
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$$200$$
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$$250/\pi$$
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$$500$$
Explanation
From the formula, $$\omega =\displaystyle\frac{1}{\sqrt{LC}}$$
Given, $$L=8$$H,
$$C=0.5\mu F=0.5\times 10^{-6}F$$
$$\therefore \omega =\displaystyle\frac{1}{\sqrt{8\times 0.5\times 10^{-6}}}$$
$$\Rightarrow \omega =\displaystyle\frac{1}{2\times 10^{-3}}=500$$ rad/s
In an $$L-C-R$$ series circuit, the values of $$R, X_{L}$$ and $$X_{C}$$ are $$120\Omega, 180\Omega$$ and $$130\Omega$$, what is the impedance of the circuit?
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$$120\Omega$$
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$$130\Omega$$
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$$180\Omega$$
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$$330\Omega$$
Explanation
Given, $$R = 120\Omega$$
$$X_{L} = 180\Omega$$ and $$ X_{C} = 130\Omega$$
The impedance of $$L-C-R$$ circuit
$$Z = \sqrt {R^{2} + (X_{L} - X_{C})^{2}}$$
$$Z = \sqrt {(120)^{2} + (180 - 130)^{2}}$$
$$Z= 130\Omega$$.
A series resonant circuit contains $$L = \dfrac{5}{\pi} mH, C = \dfrac{200}{\pi} \mu F$$ and $$R = 100 \mu$$. If a source of emf $$e = 200 sin 1000 \pi t$$ is applied, then the rms current is:
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$$2A$$
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$$200 \sqrt 2 A$$
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$$100 \sqrt 2 A$$
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$$1.41 A$$
Explanation
Source emf is given as $$e = 200 sin 1000 \pi t$$
Comparing it with $$e = e_o \sin wt$$
We get $$ \omega = 1000 \pi$$
$$\Rightarrow f = \dfrac{\omega}{2 \pi} = 500 Hz$$
Also, $$e_0 = 200 V$$
At resonance $$Z = R$$
So, current flowing $$i_0 = \dfrac{e_o}{R} = \dfrac{200}{100} = 2A$$
Rms value of current $$i_{rms} = \dfrac{i_0}{\sqrt 2} = \dfrac{2}{\sqrt 2} = \sqrt 2 = 1.41 A$$
The alternating voltage and current in an electric circuit are respectively given by $$E=100\sin { 100\pi t } ,I=5\sin { 100\pi t } $$. The reactance of the circuit will be :
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$$1\Omega $$
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$$0.05\Omega $$
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$$20\Omega $$
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Zero
Explanation
The general equations of alternating voltage and current in an AC circuit are as follows
$$E={ E }_{ 0 }\sin { \omega t } $$ ....(i)
$$I={ I }_{ 0 }\sin { \omega t } $$ ....(ii)
Given that,
$$E=100\sin { 100\pi t } $$ ....(iii)
$$I=5\sin { 100\pi t } $$ ....(iv)
Comparing equations (ii) and (iv), we get,
Peak value of current, $${ I }_{ 0 }=5A$$
Comparing equations (i) and (iii), we get
Peak value of voltage, $${ E }_{ 0 }=100 V$$
Now, r.m.s. value of voltage is given by
$${ E }_{ rms }=\dfrac { { E }_{ 0 } }{ \sqrt { 2 } } =\dfrac { 100 }{ \sqrt { 2 } } V$$
Similarly, r.m.s. value of current is given by
$${ I }_{ rms }=\dfrac { { I }_{ 0 } }{ \sqrt { 2 } } =\dfrac { 5 }{ \sqrt { 2 } } A$$
Hence, reactance of the circuit is given by
$$Z=\dfrac { { E }_{ rms } }{ { I }_{ rms } } =\dfrac { \dfrac { 100 }{ \sqrt { 2 } } }{ \dfrac { 5 }{ \sqrt { 2 } } } =20\Omega $$
In non-resonant circuit, what will be the nature of the circuit for frequencies higher than the resonant frequency?
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$$Resistive$$
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$$Capacitive$$
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$$Inductive$$
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$$None\ of\ these$$
Explanation
At resonant frequency
$$X_L=X_C$$ $$\left(\omega L=\displaystyle\frac{1}{\omega C}\right)$$
At frequencies higher than resonance frequencies
$$X_L > X_C$$
i.e., behaviour is inductive.
In a circuit, $$L,\ C$$ and $$R$$ are connected in series with an alternating voltage source of frequency $$f$$. The current leads the voltage by $$45^o$$. The value of $$C$$ is :
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$$ \dfrac {1}{\pi f ( 2 \pi f L + R)} $$
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$$ \dfrac {1}{\pi f ( 2 \pi f L - R)} $$
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$$ \dfrac {1}{2 \pi f ( 2 \pi f L + R)} $$
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$$ \dfrac {1}{2 \pi f ( 2 \pi f L - R)} $$
Explanation
The phase differenec $$ \phi $$ between current and voltage is given by
$$ \tan \phi = \dfrac {X_L - X_C}{R} $$
$$ \tan 45^o = 1 $$
$$ X_C = X_L - R $$
$$ \dfrac {1}{2 \pi f C } = 2 \pi fL - R $$
$$ C = \dfrac {1}{2 \pi f (2 \pi f L - R)} $$
A $$220$$V main supply is connected to a resistance of $$100$$k$$\Omega$$. The effective current is?
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$$2.2$$mA
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$$2.2\sqrt{2}$$mA
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$$\displaystyle\frac{2.2}{\sqrt{2}}$$mA
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None of these
Explanation
Effective current is the rms value. Here, $$220$$V is the labelled value of AC which is also the rms value. Hence,
$$I_{rms}=\displaystyle\frac{E_{rms}}{R}$$
$$I_{rms}=\displaystyle\frac{220}{100\times 10^3}$$
$$I_{rms}=2.2mA$$.
Which one of the following curves represents variation of current i with frequency f in series LCR circuit?
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0%
0%
0%
Explanation
In series LCR circuit, the graph(c) represents the curve between current (i) and frequency(f).
For the given circuit, the natural frequency is given by :
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$$\frac {1}{2\pi}\sqrt{LC}$$
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$$\frac {1}{2\pi\sqrt{LC}}$$
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$$\sqrt{\frac {L}{C}}$$
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$$\sqrt{\frac {C}{L}}$$
Explanation
Natural frequency $$(\omega)$$ is given by
$$f=\dfrac {1}{2\pi} \cdot \dfrac{1}{\sqrt{LC}}$$
For the given circuit $$C_{net}=\dfrac{C}{2}$$ (in series)
$$L_{net}=2L$$ (in series)
$$f=\dfrac {1}{2\pi} \cdot \dfrac{1}{\sqrt{LC}}$$
A 44mH inductor is connected to a 220V, 50Hz AC supply. Then, the inductive reactance of the inductor is :
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$$3.82 \, \Omega $$
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$$10.8 \, Hz$$
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$$13.82 \, \Omega $$
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$$21.5 \, Hz$$
Explanation
Inductive reactance $$X_L$$ is given by
$$X_L = \omega L = 2\pi fL$$
Given, f=50Hz, L=44mH
$$= 44\times 10^{-3}H$$
$$X_L = 2\pi \times 44 \times 10^{-3} \times 50= 13.82 \Omega$$
In the given circuit $$R$$ in pure resistance and $$X$$ is unknown circuit element. An AC voltage source is applied across $$A$$ and $$C$$. If $${ V }_{ AB }={ V }_{ AC }$$, then $$X$$ is
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Pure resistance
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Pure inductance
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Combination of inductance and capacitance at resonance
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None of the above
Explanation
Since the voltage across the resistive element is same as the voltage applied, the voltage drop across $$BC$$ is zero. This is possible only when the ohmic value of the element connected across $$BC$$ is zero. So, $$X$$ should be combination of inductance and capacitance at resonance.
The frequency of the output signal becomes ________ times by doubling the value of the capacitance in the LC oscillator circuit.
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$$\sqrt { 2 } $$
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$$\dfrac { 1 }{ \sqrt { 2 } } $$
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$$\dfrac { 1 }{ 2 } $$
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$$2$$
Explanation
In the L-C circuits
For the first condition
$${ f }_{ 1 }=\dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 1 }{ LC } } $$ .....(i)
For the second condition, when $$C=$$ doubling
$${ f }_{ 2 }=\dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 1 }{ 2LC } } $$ ....(ii)
From the equation (i), we get
$${ f }_{ 2 }=\dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 1 }{ LC } } \times \dfrac { 1 }{ \sqrt { 2 } } \Rightarrow { f }_{ 2 }={ f }_{ 1 }\times \dfrac { 1 }{ \sqrt { 2 } } $$
So, the frequency of the output signal becomes $${ 1 }/{ \sqrt { 2 } }$$ times by doubling the value of the capacitance in the L-C oscillator circuit.
In a series resonant circuit, the AC voltage across resistance R, inductor L and capacitor C are 5 V, 10 V and 10 V respectively. The AC voltage applied to the circuit will be
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10 V
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25 V
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5 V
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20 V
Explanation
Given, $$V_R=5\, V, V_L=10$$ and $$V_C=10 \, V$$
In the MR circuit, the AC voltage applied to the circuit will be
$$V=\sqrt{V^2_R+(V_L-V_C)^2}$$
$$=\sqrt{(5)^2+(10-10)^2}$$
$$=5\, V$$
The power loss in an AC circuit can be minimized by.
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Decreasing resistance and increasing inductance
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Decreasing inductance and increasimg resistance
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Increasing both inductance and resistance
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Decreasing both inductance and resistance
Explanation
In an AC circuit, power loss can be minimized by decreasing in resistance and by increasing in inductance.
The diagram given show the variation of voltage and current in an AC circuit. The circuit contains
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Only a resistor
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Only a pure inductor
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Only a capcacitor
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A capacitor and and inductor
Explanation
The given circuit shows that the current lags the applied voltage. This is possible if the circuit has inductive element. So, the circuit contain a pure inductor.
What is the range of the characteristic impedance of a coaxial cable?
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Between $$150\Omega$$ to $$600\Omega$$
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Between $$50\Omega$$ to $$70\Omega$$
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Between $$0\Omega$$ to $$50\Omega$$
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Between $$100\Omega$$ to $$150\Omega$$
Explanation
Characteristic impedance of a coaxial cable is between $$50\Omega$$ to $$70\Omega$$.
In the given circuit the potential difference across resistance is $$54$$V and power consumed by it is $$16$$W. If AC frequency is $$60$$ Hz find the value of L.
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$$1$$H
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$$2$$H
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$$5$$H
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$$4$$H
Consider the $$R-L-C$$ circuit given below. The circuit is driven by a $$50\ Hz\ AC$$ source with peak voltage $$220\ V$$. If $$R = 400\Omega, C = 200\mu F$$ and $$L = 6H$$, the maximum current in the circuit is closest to
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$$0.120\ A$$
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$$0.55\ A$$
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$$1.2\ A$$
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$$5.5\ A$$
Explanation
At maximum current $$X_{c} = X_{L}$$
so Z = R
so $$I = \dfrac{v}{R}$$
so I = 0.55 A
Consider two series resonant circuits with components $$L_1C_1$$ and $$L_2C_2$$ with same resonant frequency , $$\omega$$.When connected in series, the resonant frequency of the combination is
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$$2\omega$$
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$$\dfrac{\omega}{2}$$
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$$3\omega$$
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$$\omega$$
Explanation
Initially, $$\omega = \dfrac{1}{\sqrt{L_1C_1}}$$ and
$$\omega = \dfrac{1}{\sqrt{L_2C_2}}$$
Finally, $$L_{eq} = L_1 + L_2$$
$$C_{eq} = \dfrac{C_1C_2}{C_1 + C_2}$$
$$\omega' = \dfrac{1}{\sqrt{L_{eq}C_{eq}}}$$
from above equation, we will get $$\omega' = \omega$$
In the circuit shown in the figure, neglecting source resistance, the voltmeter and ammeter readings will respectively be
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0 V, 8 A
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150 V, A 8
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150 V, 3 A
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0 V, 3 A
Explanation
Solution:-
Given:- $$X_C=X_L$$, V=240V, R=30$$\Omega$$
Since X_L=X_C, So,
Z=$$\sqrt{R^2+{\left(X_L-X_C\right)}^2}$$=R
So, potential drop across C $$\&$$ L is zero and current is $$\rightarrow$$
i=$$\cfrac{V}{Z}=\cfrac{V}{R}=\cfrac{240}{30}=8A$$
i=8A
A $$5cm$$ long solenoid having $$10$$ ohm resistance and $$5mH$$ inductance is joined to a $$10V$$ battery. At steady state, the current through the solenoid (in ampere) will be
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$$5$$
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$$2$$
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$$1$$
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zero
Explanation
At steady state inductor behave like short circuit.
$$ i = \cfrac{v}{r} = \cfrac{10}{10} = 1A$$
Consider $$L, C, R$$ circuit as shown in figure, with a.c. source of peak value $$V$$ and angular frequency $$\omega$$. Then the peak value of current through the ac source.
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$$\dfrac {V}{\sqrt {\dfrac {1}{R^{2}} + \left (\omega L - \dfrac {1}{\omega C}\right )^{2}}}$$
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$$V \sqrt {\dfrac {1}{R^{2}} + \left (\omega C - \dfrac {1}{\omega L}\right )^{2}}$$
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$$\dfrac {V}{\sqrt {R^{2} + \left (\omega L - \dfrac {1}{\omega C}\right )^{2}}}$$
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$$\dfrac {VR\omega C}{\sqrt {\omega^{2}C^{2} + R(\omega^{2} C^{2} - 1)^{2}}}$$
Explanation
$$X_c=\cfrac 1 {j\omega c}$$
$$X_L=j\omega L$$
And $$R$$ are in parallel,
Therefore, equivalent admittance of the circuit is
$$Y=\cfrac { 1 }{ { X }_{ c } } +\cfrac { 1 }{ { X }_{ L } } +\cfrac { 1 }{ R } .$$
$$Y=\cfrac { 1 }{ R } +j(c\omega -\cfrac { 1 }{ L\omega } )$$
$$\left| Y \right| =\sqrt { \cfrac { 1 }{ { R }^{ 2 } } +({ \omega c-\cfrac { 1 }{ \omega L } ) }^{ 2 } } $$
Current $$I=VY=V\sqrt { { \cfrac { 1 }{ { R }^{ 2 } } +({ \omega c-\cfrac { 1 }{ \omega L } ) }^{ 2 } } } $$
Statement 1 : An inductance and a resistance are connected in series with an ac circuit. In this circuit the current and the potential difference across the resistance lags behind potential difference across the inductance by an angle of $$\pi/2$$.
Statement 2 : In a LR circuit voltage leads the current by phase angle which depends on the value of inductance and resistance both.
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Both statements 1 and 2 are true and statement 2 is the correct explanation of statement 1.
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Both statements 1 and 2 are true but statement 2 is not the correct explanation of statement 1.
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Statement 1 is true but statement 2 is false.
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Both statements 1 and 2 are false.
In A.C circuit having only capacitor, the current _____
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lags behind the voltage by $$\cfrac { \pi }{ 2 } $$ in phase
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leads the voltage by $$\cfrac { \pi }{ 2 } $$ in phase
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leads the voltage by $$\pi$$ in phase
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lags behind the voltage by $$\pi$$ in phase
Explanation
Current in the capacitive A.C circuit leads the voltage by $$\dfrac{\pi}{2}$$ in phase.
Correct answer is option B.
In the case of an inductor
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voltage lags the current by $$\dfrac{\pi}{2}$$.
0%
voltage leads the current by $$\dfrac{\pi}{2}$$.
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voltage leads the current by $$\dfrac{\pi}{3}$$.
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voltage leads the current by $$\dfrac{\pi}{4}$$.
Explanation
In the purely inductive circuit,
Input voltage, $$V=V_m sin (\omega t)$$ . . . . . . .(1)
The induced emf across the inductor
$$e=-L\dfrac{dI}{dt}$$
The induced emf in the circuit is equal and opposite to the applied voltage
$$e=-V$$
$$V_msin\omega t=L\dfrac{dI}{dt}$$
$$dI=\dfrac{V_m}{L}sin\omega t$$
$$I=\int dI=\dfrac{V_m}{\omega L}(-cos\omega t)$$
$$I=\dfrac{V_m}{X_L}sin(\omega t-\pi /2)$$
$$I=I_msin(\omega t-\pi /2)$$. . . . . .(2)
where, $$I_m=\dfrac{V_m}{X_L}$$
$$X_L=\omega L$$ (inductive reactants)
From equation (1) and (2)
Voltage leads the current by $$90^0$$.
The correct option is B.
Two coils have mutual inductance $$0.005H$$. The current changes in the first coil according to equation $$I={ I }_{ 0 }\sin { \omega t } $$, where $${ I }_{ 0 }=10A$$ and $$\omega =100\pi rad\quad { s }^{ -1 }$$. The maximum value of emf in the second coil is
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$$5$$
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$$5\pi $$
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$$0.5\pi $$
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$$\pi$$
Explanation
Current is given as $$I = I_o \ \sin wt = 10\sin (100\pi t)$$
Rate of change of current $$\dfrac{dI}{dt} = I_o w \cos wt = 1000\pi \cos (100\pi t)$$
Thus $$\dfrac{dI}{dt}\bigg|_{max} = 1000\pi $$
Given : $$L = 0.005 \ H$$
Maximum emf $$E_{max} = L\dfrac{dI}{dt}\bigg|_{max} = 0.005\times 1000\pi = 5\pi$$
Which of the following graphs represents the correct variation of inductive reactane $$X_L$$ with frequency $$ \upsilon$$?
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0%
0%
0%
Explanation
Inductive reactance ,
$$X_L = \omega L = 2 \pi\nu L $$
$$X_L= (2\pi L) \times\nu $$
The above equation can be compared to the equation of a straight line.
$$y = m \times x$$
Hence, inductive reactance increases linearly with frequency
An ideal inductor is in turn put across $$220 V, 50 Hz$$ and $$220 V, 100 Hz$$ supplies. The current flowing through it in the two cases will be then
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equal
0%
different
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zero
0%
infinite
Explanation
The current in the inductor coil is given by $$I \, = \, \dfrac{V}{X_L} \, = \, \dfrac{V}{2\pi \upsilon L}$$
Since frequency $$\upsilon$$ in the two cases is different, hence the current in two cases will be different.
The reciprocal of the impedance of an electric current is?
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Admittance
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Susceptance
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Conductance
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Reactance
Explanation
As conductance is the complement of resistance, there is also a complementary expression of reactance or impedance, called susceptance
A capacitor 'C' is connected across a D.C. source, the reactance of capacitor will be _________.
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ZERO
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HIGH
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LOW
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INFINITE
Explanation
Frequency of a source voltage $$f = 0$$
Reactance of capacitance $$X_C = \dfrac{1}{2\pi f C}$$
$$\implies \ X_C = \infty$$
Correct answer is option D.
A 5 $$\mu$$F capacitor is connected to a 200 V, 100 Hz ac source. The capacitive reactance is:
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$$212 \Omega$$
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$$312 \Omega$$
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$$318 \Omega$$
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$$412 \Omega$$
Explanation
Given:
The capacitance of the capacitor is $$5\mu F$$.
The voltage supplied is $$200\ V$$.
The frequency of the voltage supplied is $$100\ Hz$$.
To find:
The capacitive reactance of the capacitor.
The reactance of the capacitor is given by:
$$X_C=\cfrac{1}{\omega_C}=\cfrac{1}{2\pi f_C}$$
$$\Rightarrow\cfrac{1}{2\pi\times100\times5}\\ \quad=318\ \Omega$$
Which of the following graphs represents the correct variation of capacitive reactance $$X_C$$ with frequency $$\nu$$?
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0%
0%
0%
Explanation
Capacitive reactance ,
$$X_c = \dfrac{1}{\omega C} \, = \, \dfrac{1}{2 \pi\nu C} $$
$$ \Rightarrow X_C \times \nu= constant$$
With an increase in frequency,$$X_C$$decreases
Hence, the option (c) represents the correct graph.
An inductor of 30 mH is connected to a 220 V, 100 Hz ac source. The inductive reactance is then
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$$10.58 \Omega$$
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$$12.64 \Omega$$
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$$18.85 \Omega$$
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$$22.67 \Omega$$
Explanation
Hence ,$$ L = 30 mH = 30 \, \times \, 10^{-3} \, H$$
$$V_{rms} \, = \, 220, \, \upsilon \, = \, 100 \, Hz$$
Inductive reactance,
$$X_L \, = \, 2 \pi \upsilon L \, = \, 2 \, \times \, 3.14 \, \times \, 100 \, \times \, 30 \, \times \, 10^{-3} \, = \, 18.85 \, \Omega$$
In a series LCR circuit, the plot of $$I_m$$ vs $$ \omega$$ is shown in the figure. The bandwidth of this plot will be then
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0%
Zero
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$$0.1 rad s^{-1}$$
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$$0.2 rad s^{-1}$$
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$$0.4 rad s^{-1}$$
Explanation
Bandwidth is the frequency range at which the current amplitude is
$$\dfrac{1}{\sqrt2}$$ times the magnitude of the current.
$$I_m \, = \, \dfrac{1}{\sqrt{2}} \, I_{max} \, = \, 0.7 I_{max}.$$
From figure,
Band width at this value is :
$$\Delta \omega \, = \, 1.2 \, - \, 0.8 \, = \, 0.4 \, rad \, s^{-1}$$
In the series LCR circuit shown the impedance is:
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$$200 \Omega$$
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$$100\Omega$$
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$$300\Omega$$
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$$500\Omega$$
Explanation
Hence , $$L = 1 H , C = 20 \mu F = 20 \times 10^{-6}F$$
R$$ = 300 \Omega, \upsilon = \dfrac{50}{\pi} Hz$$
The inductive reactance is
$$X_L = 2 \pi \upsilon L \, = \, 2 \times \pi \times \dfrac{50}{\pi} \times 1 = 100 \Omega$$
The capacitive reactance is
$$X_C = \dfrac{1}{2 \pi \upsilon C} = \dfrac{1}{2 \times \pi \times \dfrac{50}{\pi} \times 20 \times 10^{-6}} = 500 \Omega$$
The impedance of the series LCR cicuit is
$$ Z = \sqrt{R^2 + (X_C -X_L)^2}\,\, = \,\,\sqrt{(300)^2 + (500-100)^2} $$
$$Z = \sqrt{(300)^2 + (400)^2} = 500\, \Omega $$
In an alternating current circuit consisting of elements in series, the current increases on increasing the frequency of supply. Which of the following elements are likely to constitute the circuit?
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Only resistor
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Resistor and inductor
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Resistor and capacitor
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Only a capacitor
Explanation
We know,
Resistance (reactance) of the inductor and capacitor depends upon the frequency of the current.
$$X_c=\dfrac{1}{2\pi\ fC}$$
$$X_l=2\pi fL$$
$$I_{RMS}=\dfrac{V_{RMS}}{\sqrt{R^2+(X_L-X_C)^2}}$$
The RMS current for a resistor will not be affected because its impedance is independent of frequency whereas in the case of a resistor and inductor also, the current will decrease with an increase in frequency.
whereas the increase in frequency in the case of the capacitor leads to an increase in the RMS current because in this the impedance decreases with an increase in frequency.
Hence Option
$$\textbf C$$ is correct answer.
Phase difference between voltage and current in a capacitor in an ac circuit is then
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$$\pi$$
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$$\pi/2$$
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$$0$$
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$$\pi/3$$
Explanation
In a capacitive ac circuits , the voltage lags
behind the current in phase by $$\pi /2$$ radian.
A circuit containing a 20 $$\Omega$$ resistor and 0.1$$\mu F$$ capacitor in series is connected to 230 V AC supply of angular frequency 100 rad $$s^{-1}$$. The impedance of the circuit is
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$$10^5\Omega$$
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$$10^4\Omega$$
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$$10^6\Omega$$
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$$10^{10}\Omega$$
Explanation
Hence, $$R = 20 \Omega, C = 0.1 \mu F= 0.1 \times 10^{-6}F = 10^{-7}F$$
Impedance ,$$ Z = \sqrt{R^2 + \dfrac{1}{\omega^2C^2}}$$
$$\sqrt{20^2 + \dfrac{1}{(100)^2 \times (10^{-7})^2}} \, = \, \sqrt{400 + 10^{10}} = 10^5 \, \Omega$$
A sinusoidal voltage of peak value 293 V and frequency 50 Hz is applied to a series LCR circuit in which R = 6 $$\Omega$$, L = 25 mH and C = 750 $$\mu$$F. The impedance of the circuit is:
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7.0
$$\Omega$$
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8.9
$$\Omega$$
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9.9
$$\Omega$$
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10.0
$$\Omega$$
Explanation
Hence , $$R = 6\Omega, L = 25 mH = 25 \times10^{-3}H$$
C$$ = 750\mu F= 750\times10^{-6}F, \upsilon = 50 Hz$$
$$X_L = 2\pi \upsilon L = 2\times3.14\times50\times25\times10^{-3} = 7.85 \Omega$$
$$X_C = \dfrac{1}{2\pi \upsilon C} = \dfrac{1}{2 \times 3.14 \times 50\times 750 \times 10^{-6}}= 4.25 \Omega$$
$$\therefore X_L -X_C = 7.85 - 4.25 = 3.6\Omega$$
Impedence of the series LCR circuit is
$$Z = \sqrt {R^2 + ( X_L - X_C)^2}$$
$$\therefore Z = \sqrt{(6)^2+(3.6)^2} = \sqrt{36 + 12.96} = 7.0\Omega$$
The resonant frequency of a series LCR circuit with L = 2.0H,C=32$$\mu$$F and R= 10 $$\Omega$$ is
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$$20 Hz$$
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$$30 Hz$$
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$$40 Hz$$
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$$50 Hz$$
Explanation
Here, $$L = 2H$$
$$C = 32\mu F = 32 \times 10^{-6}F $$
$$ R = 10\Omega $$
$$\therefore \omega = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{2\times32\times10{-6}}} =125\, rad s^{-1}$$
$$\nu_r = \dfrac{\omega_r}{2\pi} = \dfrac{125}{2\times3.14}= 20Hz$$
A 0.2 $$\Omega$$ resistor and 15 $$\mu$$F capacitor are connected in series to a 220 V, 50 Hz ac source. The impedance of the circuit is
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$$250 \Omega$$
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$$268 \Omega$$
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$$29.15 \Omega$$
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$$291.5 \Omega$$
Explanation
Here, $$R = 0.2 k \Omega = 200 \Omega$$
$$C = 15 \mu F = 15 \times 10^{-6}F$$
$$\nu= 50Hz$$
Capacitive reactance,
$$ X_C = \dfrac{1}{2 \pi \nu C} = \dfrac{1}{2 \times 3.14 \times 50 \times 15 \times 10^{-6}} = 212 \, \Omega$$
The impedance of the RC circuit is
$$ Z = \sqrt{R^2 + X^2_C} = \sqrt{(200)^2 + (212)^2} = 291.5 \, \Omega$$
A coil of inductance $$L = 300mH$$ and resistance $$R = 140 m\Omega$$ is connected to a constant voltage source, Current in the coil will reach to $$50%$$ of its steady state value after $$t$$ is nearly equal to:
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$$15 s$$
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$$0.75 s$$
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$$0.15 s$$
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$$1.5 s$$
Explanation
$$\dfrac{V}{2R} = \dfrac{V}{R} (1 - e^{-Rt/L})$$
$$\ell^{-\frac{Rt}{L}} = \dfrac{1}{2}$$
$$\dfrac{R}{L}t = \ell n2$$
$$t = \dfrac{300}{140} \times {0.7}{10} = 1.5$$
In a series resonant R-L-C circuit, if L is increased by $$25 \%\ $$ and C is decreased by $$20\%\ $$, then the resonant frequency will
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Increases by $$10 \%\ $$
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Decreases by $$10 \%\ $$
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Remain unchanged
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Increases by $$2.5 \%\ $$
A series resonant LCR circuit has a quality factor (Q-factor) = 0.If R 2 k$$\Omega$$, C = 0.1 $$\mu$$F, then the value of Inductance is
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0.1 H
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0.064 H
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2 H
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5 H
Explanation
Quality factor $$Q \, = \, \dfrac{1}{R} \, \sqrt{\dfrac{L}{C}} \, or \, \dfrac{L}{C} \, = \, (QR)^2$$
Here, $$Q \, = \, 0.4, \, R \, = \, 2 \, K \Omega \, = \, 2 \, \times \, 10^3 \, \omega$$
$$C \, = \, 0.1 \, \mu F \, = \, 0.1 \, \times \, 10^{-6} \, F$$
$$\therefore \, L \, = \, (QR)^2 \, C$$
$$\therefore \, L \, = \, (0.4 \, \times \, 2 \, \times \, 10^3)^2 \, \times \, 0.1 \, \times \, 10^{-6} \, = \, 0.064 \, H$$
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Practice Class 12 Medical Physics Quiz Questions and Answers
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