MCQ Questions for Class 12 Medical Physics Alternating Current Quiz 13

The value of resistance of the coil calculated by the student is :

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$3\Omega$
0%
$4\Omega$
0%
$5\Omega$
0%
$8\Omega$

Explanation

$\displaystyle V_{DC}=I_{DC}R$

$\displaystyle \therefore R=\frac{V_{DC}}{I_{DC}}=\frac{12}{4}=3\Omega$

$\displaystyle I_{AC}=\frac{V_{AC}}{Z}=\frac{V_{AC}}{\sqrt {R^2+X^2_L}}$

$\displaystyle 2.4=\frac{12}{\sqrt {(3)^2+X^2_L}}$
Solving this equaion we get,

$\displaystyle X_L=4\Omega$

$\displaystyle X_C=\frac{1}{\omega C}=\frac{1}{50\times 2500\times 10^{-6}}$

$=8\Omega$

$\displaystyle Z=\sqrt{R^2+(X_C-X_L)^2}=5\Omega$

$\displaystyle \therefore I=\frac{V_{DC}}{Z}=\frac{12}{5}=2.4A=I_{rms}$

$\displaystyle P=I^2_{rms}R=(2.4)^2(3)$

$\displaystyle =17.28W$
At

given frequency

$X_C > X_L$ . If

$\omega$ is further decreased,

$X_C$ will increase

$\left( as X_C\propto \frac{1}{\omega}\right)$

and

$X_L$ will increase

$(as X_L \propto \omega)$ .
Therefore

$X_C-X_L$ and hence

$Z$ will increase. So, current will decrease.

In the circuit shown in figure,

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$V_R=80V$
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$X_C=50\Omega$
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$V_L=40V$
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$V_0=100V$

Explanation

(a)

$\displaystyle V_R=IR=80V$
(b)

$\displaystyle X_C=\frac{V_C}{I}=\frac{100}{2}=50V$
(c)

$\displaystyle V_L=IX_L=40V$
(d)

$\displaystyle V=V_{rms}=\sqrt{V^2_R+(V_C-V_L)^2}$

$=\displaystyle \sqrt{(80)^2+(100-40)^2}$

$=60V$

$\therefore \displaystyle V_0=\sqrt 2V_{rms}=60\sqrt 2V$

A resistor and a capacitor are connected to an A.C. supply of

$200\space V, \space 50\space Hz$ in series. The current in the circuit is

$2\space A$ . If the power consumed in the circuit is

$100\space W$ then the capacitive reactance in the circuit is

0%
$100\space \Omega$
0%
$25\space \Omega$
0%
$\sqrt{125\times75}\space \Omega$
0%
$400\space \Omega$

Explanation

$V_{rms} = 200 V$

$I_{rms} = 2A$

$P = 100 = I_{rms}^2 R = 4 R$

$R=25 \Omega$

$\therefore X_C= \sqrt { R^2 + X_C^2 } = \dfrac {V_{rms}}{I_{rms}} =100$

$\Rightarrow X_C= \sqrt{ 100^2 - 25^2 } = \sqrt{ 9375} = \sqrt{ 125 \times 75} \Omega$

$50\ Hz$ a.c. source of

$20\ V$ is connected across R and C as shown in figure below. The voltage across R is

$12\ V$ . The voltage across C is:

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$8\ V$
0%
$16\ V$
0%
$10\ V$
0%
not possible to determine unless values of R and C are given

Explanation

For ac C- R circuit,

$V_S^2=V_R^2+V_C^2$
or

$V_C=\sqrt{V_S^2-V_R^2}$

$=\sqrt{20^2-12^2}$

$=\sqrt{400-144}$

$=\sqrt{256}=16\ V$

A series

$LCR$ circuit containing a resistance of

$120\space \Omega$ has angular resonance frequency

$4\times10^5\space rad\space s^{-1}$ . At resonance the voltages across resistance and inductance are

$60\space V$ and

$40\space V$ , respectively. The value of inductance

$L$ is

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$0.1\space mH$
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$0.2\space mH$
0%
$0.35\space mH$
0%
$0.4\space mH$

Explanation

At resonance the reactance of inductor and the capaciotr cancel each other,

$L\omega = \frac{1}{C\omega}$ and total

$Z= R \Omega$
resonance frequency =

${\omega}_r = 4 \times 10^5 rad/s$
let the current at resonance be

$I$ .

$V_R = I \times R = 60 V \Rightarrow I= \frac{60}{120} A$

$V_L = I \times L{\omega}_r = 40 V \Rightarrow L=0.2 mH$

The inductance of the coil is

0%
$0.02\space H$
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$0.04\space H$
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$0.08\space H$
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$1.0\space H$

Explanation

impedance of coil=

$R +jL\omega$ .
when DC is applied,

$I_{dc}= \frac{V_{dc}}{R} \Rightarrow 4 = \frac{12}{R} \Rightarrow R= 3\Omega$
when AC(12 V , 50rad/s ) is applied,

$I_{ac} = \frac{ V_{ac}}{ \sqrt{ R^2+{(L \omega )}^2 } } \Rightarrow L = \sqrt{ \frac{1}{{\omega}^2}(\frac{V_{ac}^2}{I_{ac}^2} - R^2) } = \frac{2}{25} = .08 H$

The impedance of

$P$ at this frequency is :

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$77\space \Omega$
0%
$36\space \Omega$
0%
$40\space \Omega$
0%
$125\space \Omega$

Explanation

P has

$R_1= 32 \Omega$

$C_1=1 \mu F$
Q has

$R_2=68 \Omega$

$L_2 = 4.8 m H$
Total impedance of the series combination of P and Q :

$Z= R_1+R_2 +j L\omega -j\frac{1}{C\omega}$
The frequency is adjusted so that maximum current flows in P and Q. that is, frequency is adjusted to have minimum total series impedance. Total series impedance will be minimum when the reactance of L and C cancel each other. Therefore,

$\omega = \frac{1}{\sqrt{LC}} =\frac{100000}{7} rad/s$
Therefore at maximum current, impedance of P=

$R_1 + -j\frac{1}{C\omega} = 32 -70j =77 \Omega$

When

$100\space V$ dc is applied across a coil, a current of

$1\space A$ flows through it and when

$100\space V$ ac of

$50\space Hz$ is applied to the same coil, only

$0.5\space A$ flows. The inductance of coil is

0%
$5.5\space H$
0%
$3/\pi\space H$
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$\sqrt3/\pi\space H$
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$2.5\space H$

Explanation

$R = \frac{V_{DC}}{I_{DC} } = \frac{100}{1} = 100 \Omega$

$\frac{V_{AC}^2}{I_{AC}^2} = R^2 + X_L^2 \Rightarrow \frac{100^2}{0.5^2}= 100^2+ X_L^2 \Rightarrow X_L = \sqrt{30000}= 100\sqrt{3} \Omega$

$X_L= L 2 \pi f \Rightarrow 100\sqrt{3} = L 2 \pi 50 \Rightarrow L =\frac{\sqrt{3}}{\pi} H$

The impedance of

$Q$ at this frequency is:

0%
$200\space \Omega$
0%
$\sqrt{1350}\space \Omega$
0%
$55\space \Omega$
0%
$\sqrt{9524}\space \Omega$

Explanation

P has

$R_1= 32 \Omega$

$C_1=1 \mu F$
Q has

$R_2=68 \Omega$

$L_2 = 4.9 m H$
Total impedance of the series combination of P and Q :

$\omega = \frac{1}{\sqrt{LC}} =\frac{100000}{7} rad/s$
Therefore at maximum current, impedance of

$Q= R_2 + jL\omega = 68+70j =\sqrt{9524} \Omega$

In the circuit shown in the figure, the A.C. source gives a voltage V = 20 cos (2000 t) volt. Neglecting source resistance, select correct alternative(s).

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The reading of voltmeter is 0 V
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The reading of voltmeter is 5.6 V
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The reading of ammeter is 1.4 A
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The reading of ammeter is 0.47 A

Explanation

Here,

$V_P=20$ volt and

$\omega=2000$ rad/s

At resonance condition, the current in the circuit

$I=\dfrac{V_{rms}}{R}=\dfrac{V_P/\sqrt2}{6+4}=$

In a purely inductive circuit, the current is

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In phase with the voltage
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Out of phase with the voltage
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Leads the voltage by $\pi/ 2$
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Lags behind the voltage $\pi/ 2$

If the power factor in a circuit is unit, then the impedance of the circuit is

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Inductive
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Capacitive
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Partially inductive and partially capacitive
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Resistive

An oscillating circuit of a capacitor with capacitance

$C$ , a coil of inductance

$L$ with negligible resistance, and switch. With the switch disconnected the capacitor was charged to a voltage

$V_m$ and then at the moment

$t=0$ , the switch was closed. The current

$I$ in the circuit as a function of time is represented as

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$V_m\sqrt{\frac{L}{C}} sin (wt)$
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$V_m\sqrt{\frac{C}{L}} sin (wt)$
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$V_m\sqrt{\frac{L}{C}} cos (wt)$
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$V_m\sqrt{\frac{C}{L}} sin (wt)$

A square conducting loop of side L is situated in gravity free space. A small conducting circular loop of radius r ( r < < L) is placed at the center of the square loop, with its plane perpendicular to the plane of the square loop. The mutual inductance of the two coils is

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$\dfrac{2\sqrt2 \mu_sI}{L}r^2$
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$\dfrac{\sqrt2 \mu_sI}{L}r^2$
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0
0%
None of these

In a LR circuit of

$3\;mH$ inductane and

$4\;\Omega$ resistance, emf

$E=4\;cos\;(1000\;t)$ volt is applied. The amplitude of current is:

0%
$0.8\ {A}$
0%
$\dfrac{4}{7}\ {A}$
0%
$1.0\ {A}$
0%
$\dfrac{4}{\sqrt{7}}\ {A}$

Explanation

Given :

$L = 3mH = 3 \times 10^{-3} H$

$R =4 \Omega$

Emf

$E = 4 cos1000t$

$\therefore$

$E_o = 4$ volt and

$\omega = 1000$

Inductive reactance

$X_L = \omega L = 1000 \times 3 \times 10^{-3} = 3 \Omega$

Impedance

$Z = \sqrt{R^2 + X_L^2} = \sqrt{4^2 + 3^2} = 5 \Omega$

Current amplitude

$I= \dfrac{E_o}{Z} = \dfrac{4}{5} = 0.8 A$

A series LCR circuit is connected across a source of alternating emf of changing frequency and resonates at frequency

$\text{f}_o$ . Keeping capacitance constant, if the inductance (L) is increased by

$\sqrt{3}$ times and resistance(R) is increased by 1.4 times, the resonant frequency now is:

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$3^{1/4}\text{f}_0$
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$\sqrt{3}\text{f}_0$
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$(\sqrt{3}-1)^{1/4}\text{f}_0$
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$\left(\dfrac{1}{3}\right)^{1/4}\text{f}_0$

Explanation

Resonant frequency of a series LCR circuit is given by the formula

$f=\dfrac { 1 }{ 2\pi \sqrt { LC } }$

Hence,

${ f }_{ 0 }=\dfrac { 1 }{ 2\pi \sqrt { LC } }$

In second case,

${ f }'=\dfrac { 1 }{ 2\pi \sqrt { L'C } } =\dfrac { 1 }{ 2\pi \sqrt { \sqrt { 3 } LC } } =\dfrac { 1 }{ { 3 }^{ \frac { 1 }{ 4 } } } { f }_{ 0 }$

The given graph shows variation with time in the source voltage and steady state current drawn by a series RLC circuit.
Which of the following statements is/are correct?

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Current lags the voltage
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Resistance in the circuit is 250 $\sqrt {3} \Omega$
0%
Reactance in the circuit is 250 $\Omega$
0%
Average power dissipation in the circuit is 20 $\sqrt {3}$ W

Explanation

For option (A) Solution :

Expression of voltage of time t:

$v= v_m$ (

$\sin \omega t + \phi$ )

Expression of current of time t:

$i = i_m$ (

$\sin \omega t$ )

At t = 0 ;

$100 = 200\sin \displaystyle \phi \Rightarrow \phi = \frac {\pi} {6}$ rad ; current lags voltage by

$\displaystyle \frac {\pi} {6}$ rad

For option (B) Solution :

$R=Z \cos \displaystyle \phi = \frac{V}{I}\cos \phi= \frac {200V} {400mA}\,cos \frac {\pi} {6} = 250 \sqrt {3} \Omega$

For option (C) Solution :

$\displaystyle X=Z\sin \phi = \frac{V}{I}\sin \phi=\frac{200}{400}\sin(\pi/6) =$

$250\Omega$
For option (D) Solution :

$\displaystyle P_{av} = \frac {V_m I_m} {2}\,cos\,\phi = 20 \sqrt {3} W$

A charged capacitor discharges through a resistance R with time constant

$\tau$ . The two are now placed in series across an AC source of angular frequency

$\displaystyle \omega = \frac {1} {\tau}$ . The impedance of the circuit will be:

0%
$\displaystyle \frac {R} {\sqrt {2}}$
0%
R
0%
$\sqrt {2}$ R
0%
2R

Explanation

In RC circuit, impedance

$Z=\sqrt{R^2+\dfrac{1}{\omega^2C^2}}$ and time constant

$\tau=CR$

Since

$\omega=\dfrac{1}{\tau}$ so

$Z=\sqrt{R^2+\dfrac{\tau^2}{C^2}}=\sqrt{R^2+\dfrac{C^2R^2}{C^2}}=\sqrt{2} R$

An AC generator producing

$10V$ (rms) at

$200rad/s$ is connected in series with a

$50\Omega$ resistor, a

$400mH$ inductor and a

$200\mu F$ capacitor. The rms voltage across the inductor is

0%
$2.5V$
0%
$3.4V$
0%
$6.7V$
0%
$10.8V$

Explanation

Given :

$E=10V, \ \ \ w=200, \ \ \ \ R=50\Omega , \ \ \ \ L=400mH, \ \ \ C=200\mu F\quad$
So, capacitive reactance

$X_C = \dfrac{1}{wC} = \dfrac{1}{(200)(200\times 10^{-6})} = 25\Omega$
Inductive reactance

$X_L = wL =(200)(400\times 10^{-3}) =80\Omega$
Impedance

$\quad Z=\sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 } } =\sqrt { { 50 }^{ 2 }+{ (80-25) }^{ 2 } }$

$\quad Z=74.3\Omega$

$I=\cfrac { E }{ Z } =\cfrac { 10 }{ 74.3 } =0.13459A$

${ E }_{ L }=I{ X }_{ L }=0.1345\times 80=10.76V$

$200$ km long telegraph wire has a capacitance of

$0.014$

$\mu F/km$ . If it carries an alternating current of

$50\times 10^3$ Hz, what should be the value of an inductance required to be connected in series so that impedance is minimum?

0%
$0.48\times 10^{-2}$ mH
0%
$0.36\times 10^{-2}$ mH
0%
$0.52\times 10^{-2}$ mH
0%
$0.49\times 10^{-2}$ mH

Explanation

Given,

$C'=0.014\mu F/km$

$f=50\times 10^3$ Hz
Total capacitance,

$C=C'\times 200$

$=0.014\mu F\times 200=2.8\mu F$
For impedance to be minimum

$X_L=X_C,$ i.e.,

$\omega L=\displaystyle\frac{1}{\omega C}$

$\displaystyle L=\frac{1}{\omega ^2C}$

$=\displaystyle\frac{1}{4\pi^2l^2C}$

$=\displaystyle\frac{1}{4\pi^2\times (50\times 10^3)^2\times 2.8\times 10^{-6}}$

$=0.36\times 10^{-3}H=0.36\times 10^{-2}$ mH.

An inductor and a resistor are connected to an ac supply of

$50 V$ and

$50 Hz$ . If the voltage across the resistor is

$40 V$ the voltage across the inductor will be:

0%
$10\ V$
0%
$20\ V$
0%
$30\ V$
0%
$60\ V$

Explanation

Given: Supply voltage,

$V_S=50\ V$

Voltage across resistor,

$V_R=40\ V$

For an LR circuit:

Applied voltage

$V_S=\sqrt { { V }_{ R }^{ 2 }+{ V }_{ L }^{ 2 } }$

$\Rightarrow { V }_{ L }=\sqrt { { 50 }^{ 2 }-{ 40 }^{ 2 } } =30\ V$

In an L-R circuit, the voltage is given by

$V$ as

$283 sin 314t$ . The current is found tc be

$4 \, sin \left ( 314t-\dfrac {\pi}{4} \right )$ Calculate the resistance of the circuit.

0%
$90\Omega$
0%
$25 \Omega$
0%
$50\Omega$
0%
$75\Omega$

Explanation

In L-R series circuit current lags the voltage by an angle,

$\phi =tan^{-1}\left ( \dfrac{X_L}{R} \right )$

$\phi =\dfrac{\pi}{4}, X_L=R, \omega L=R$
Given,

$V= 283 \, sin \, 314t$
Comparing with standard equation

$V=V_0 sin\, \omega\, t$

$\omega =314$

$314 L =R$
Further

$V_0 =i_0 |Z|$

$283 = 4\sqrt{R^2+X^2_L}$

$R^2 +(\omega L)^2=\left ( \dfrac{283}{4} \right )^2=5005.86$

$2R^2 = 5005.56$

$R= 50 \Omega$

A RC series circuit of R

$=15\Omega$ and

$C=10\mu F$ is connected to

$20$ volt DC supply for very long time. Then capacitor is disconnected from circuit and connected to inductor of

$10$ mH. Find amplitude of current.

0%
$0.2\sqrt{10}$ A
0%
$2\sqrt{10}$ A
0%
$0.2$ A
0%
$\sqrt{10}$ A

Explanation

Given :

$C = 10\mu F$

$V_o = 20$ volts

$L = 10mH$
Amplitude of the current

$I_o=Q_o w=(CV_o)\dfrac{1}{\sqrt{LC}} = \sqrt{\dfrac{C}{L}}V_o$

$\implies \$

$I_o = \sqrt{\dfrac{10\times 10^{-60}}{10\times 10^{-3}}}\times 20 = 0.2\sqrt{10} A$

An a.c. source of angular frequency w is fed across a resistor R and a capacitor C in series. It registers a certain current. The frequency of the source decreases by two-third of the original value, maintaining the same voltage, the current in the circuit is found to be halved. Find the ratio of reactance to resistance at the original frequency

0%
$\sqrt{\frac{3}{5}}$
0%
$\sqrt{\frac{5}{3}}$
0%
$\sqrt{\frac{2}{3}}$
0%
$\sqrt{\frac{3}{2}}$

An L-C-R series circuit containing a resistance of

$R = 120 \Omega$ has angular resonant frequency

$4 \times 10^5$ rad/s. At resonance the voltage across resistance and inductance are 60 V and 90 V respectively. Then values of L and C are :

0%
$0.2 \, mH, \dfrac{1}{16}\mu F$
0%
$0.2 \, mH, \dfrac{1}{32}\mu F$
0%
$0.4 \, mH, \dfrac{1}{32}\mu F$
0%
$0.4 \, mH, \dfrac{1}{16}\mu F$

Explanation

$R= 120 \Omega , \omega _r=4 \times 10^5$ rad/sec

$V_R=60V, V_L=40V$

$V_R=iR \Rightarrow 60 = i \times 120$

$i= \dfrac{1}{2}A$

$V_L=iX_L\Rightarrow 40=\dfrac{1}{2}X_L$

$\therefore X_L=80 \Omega$

$\omega _rL=80\Rightarrow 4 \times 10^5 L=80$

$L=20 \times 10^{-5}H=0.2 mH$

$X_C=X_L=80 \Rightarrow \dfrac{1}{\omega _rC}=80$

$C=\dfrac{1}{80\omega }=\dfrac{1}{80\times 4\times 10^5}$

$\dfrac{10^{-6}}{32}=\dfrac{1}{32}\mu F$

In the circuit diagram shown,

$X_C=100 \Omega,X_L=200 \Omega\, and \, R=100 \Omega$ . Effective current through the source is :

0%
$2A$
0%
$2\sqrt{2}A$
0%
$0.5 A$
0%
$\sqrt {0.4} A$

The equivalent inductance between

$A$ and

$B$ is:

0%
$1 H$
0%
$4 H$
0%
$0.8 H$
0%
$16 H$

Explanation

Here, all the inductances are connected in parallel.
Hence, the equivalent inductance between A and B is

$\displaystyle \frac{1}{L_{AB}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{4}{4} = 1$
or

$L_{AB} = 1 H$
1071055_949843_ans_ee3d6c3dd6724eb5b8f78121c4ef1199.png

Two parallel wires in the plane of the paper are distance

$X_0$ apart. A point charge is moving with speed u between the wires in the same plane at a distance

$X_1$ from one of the wires. When the wires carry current of magnitude I in the same direction, the radius of curvature of the path of the point charge is

$R_1$ . In contract, if the currents I in the two wires have directions opposite to each other, the radius of curvature of the path is

$R_2$ . If

$\dfrac{X_0}{X_1}=3$ , the value of

$\dfrac{R_1}{R_2}$ is?

0%
6
0%
$3$ .
0%
4
0%
5

An alternating voltage given as

$V=100\sqrt { 2 } \sin { 100t } \quad$ is applied to a capacitor of

$1\mu F$ . The current reading of the ammeter will be equal to ______ mA.

0%
$10$
0%
$20$
0%
$40$
0%
$80$

Explanation

Given :

$C = 1\mu F$
On comparing the given voltage with

$V = V_o \ \sin wt$
we get

$V_o =100\sqrt{2}$ and

$w = 100$
Capacitive reactance

$X_C = \dfrac{1}{wC} = \dfrac{1}{100\times 10^{-6}} = 10^4\Omega$
Rms value of voltage

$V_{rms} = \dfrac{V_o}{\sqrt{2}} = 100$ volts
Thus reading of ammeter

$I_{rms} = \dfrac{V_{rms}}{I_{rms}} = \dfrac{100}{10^4} = 0.01 \ A = 10 \ mA$

In an oscillating

$LC$ circuit the maximum charge on the capacitor is

$Q$ . The charge on the capacitor when the energy is stored equally between the electric and magnetic field is

0%
$Q/2$
0%
$Q/\surd 3$
0%
$Q/\surd 2$
0%
$Q$

Explanation

[C] When capacitor is full charge, total energy of LC circuit is with capacitor

$\begin{array}{l} E=\dfrac { 1 }{ 2 } \, \, \dfrac { { { Q^{ 2 } } } }{ C } ,Q \to max\, \, { { charge } }\, \, on\, \, capacitor \\ max\, \, energy\, \, in\, \, Inductor=\dfrac { { LI_{ 0 }^{ 2 } } }{ 2 } ,{ I_{ 0 } }=current \end{array}$

According to the question

Let q charge on capacitor when charge in equally shared between inductor and capacitor.

$\begin{array}{l} \therefore \dfrac { 1 }{ 2 } \left( { \dfrac { 1 }{ 2 } \dfrac { { { Q ^{ 2 } } } }{ C } } \right) =\dfrac { 1 }{ 2 } \dfrac { { { q^{ 2 } } } }{ C } \\ \Rightarrow { Q^{ 2 } }=2{ q^{ 2 } } \\ \therefore q=\dfrac { Q}{ { \sqrt { 2 } } } \end{array}$

An electrical device draws

$2 kW$ power from ac mains voltage

$223 V(rms).$ The current differs lags in phase by

$\phi = tan^{-1} \left ( -\dfrac{3}{4} \right )$ as compared to voltage. The resistance R in the circuit is:

0%
$15\Omega$
0%
$20 \Omega$
0%
$25 \Omega$
0%
$30 \Omega$

Explanation

Here,

$P \, = \, 2 \, kW \, = \, 2 \, \times \, 10^3W$

$V_{rms} \, = \, 233 \, V, tan \, \phi \, = \, -\dfrac{3}{4}$

$As, \, P \, = \, \dfrac{V^2_{rms}}{Z}$

$\Rightarrow \, Z \, = \, \dfrac{V^2_{rms}}{P} \, = \, \dfrac{(223)^2}{2000} \, = \dfrac{49729}{2000} \, = \, 24.86 \, \Omega \, or \, Z \, = \, 25 \, \Omega$

$\tan \, \phi \, = \, \dfrac{X_C \, - \, X_L}{R} \, = \, - \dfrac{3}{4} \, \therefore \, X_C \, - \, X_L \, = \, -\dfrac{3}{4} R.$

AS,

$Z^2 \, = \, R^2 \, + \, (X_C \, - \, X_L)^2$

$\therefore \, (25)^2 \, = \, R^2 \, + \, \left(-\dfrac{3}{4} \, R \right)^2$

$625 \, = \, \dfrac{25 \, R^2}{16}.$

$R^2 \, = \, \dfrac{625 \, \times \, 16}{25} \, \, \Rightarrow \, R \, = \, 20 \, \Omega$

A coil of inductance

$300\ mH$ and resistance

$2\Omega$ is connected to a source of voltage

$2V$ . The current reaches half of its steady state value in

0%
$0.05\ s$
0%
$0.1\ s$
0%
$0.15\ s$
0%
$0.3\ s$

Explanation

Given,

Inductance,

$L=300mH=0.3H$

Resistance,

$R=2\Omega$

Source voltage,

$V_0=2V$

$I=\dfrac{I_0}{2}$

The corrent flowing in the

$LR$ circuit,

$I=I_0(1-e^{-Rt/L})$

$\dfrac{I_0}{2}=I_0(1-e^{-Rt/L})$

$1-e^{-Rt/L}=\dfrac{1}{2}$

$e^{-Rt/L}=\dfrac{1}{2}$

$\dfrac{Rt}{L}=ln2$

$t=\dfrac{Lln2}{R}=\dfrac{ln (2)\times 0.3}{2}$

$t=0.1sec$

The correct option is B.

For the circuit shown in Fig. the voltage of the source at any instant is equal to

0%
The sum of the maximum voltages across the elements.
0%
The voltage drop across the resistor.
0%
The sum of the instantaneous voltages across the elements.
0%
The sum of the rms voltages across the elements.

Explanation

The resulting velocity in an LCR circuit is the sum of the velocities in inductance and capacitance then adding the remaining voltage with that of the resistance voltage.

The angle between the voltage source and current $i$ gives the circuit phase angle.

The voltage triangle for the LCR circuit is given as

${V_s} = \sqrt {V_R^2 + {{\left( {{V_L} - {V_c}} \right)}^2}}$

$100\ volt$

$AC$ source of angular frequancy

$500\ rad/s$ is connected to a

$LCR$ circuit with

$L=0.8\ H$ ,

$C=5\ \mu F$ and

$R=10\Omega$ , all connected in series. The potential difference across the resistance is

0%
$\dfrac {100}{\sqrt {2}}volt$
0%
$100\ volt$
0%
$50\ volt$
0%
$50\sqrt {3}$

Explanation

The impedance of LCR circuit is given as,

$Z = \sqrt {\left[ {{R^2} + {{\left( {\omega L - \frac{1}{{\omega C}}} \right)}^2}} \right]}$

$= \sqrt {{{\left( {10} \right)}^2} + \left( {\left( {2 \times \pi \times 500 \times 8.1 \times {{10}^{ - 3}}} \right) - \dfrac{1}{{2 \times \pi \times 500 \times 12.5 \times {{10}^{ - 6}}}}} \right)}$

$= 10\;{\rm{ohm}}$

The rms value of the current is given as,

${I_{rms}} = \dfrac{{{V_{rms}}}}{Z}$

$= \frac{{100}}{{10}}$

$= 10\;{\rm{A}}$

The potential difference across the given resistor is given as,

${V_R} = I \times R$

$= 10 \times 10$

$= 100\;{\rm{V}}$

Thus, the potential difference across the resistor is $100\;{\rm{V}}$ .

When

$100\ V\ d.c.$ flows in a solenoid, the steady state current is

$10\ A$ . When

$100-V.\ ac$ . flows, the current drops to

$0.5\ A$ . If the frequency of ac. be

$50\ Hz$ , then the impedance and the inductance of the solenoid are.

0%
$200\Omega, 0.64\ H$
0%
$100\Omega, 0.86\ H$
0%
$200\Omega, 1.0\ H$
0%
$100\Omega, 0.93\ H$

Explanation

Resistance of solenoid

$R = \dfrac{V_{dc}}{I_{dc}} = \dfrac{100}{10} = 10\Omega$

When

$100 \ V$ AC is applied, current becomes

$0.5 \ A$ i.e.

$I_{ac} = 0.5 \ A$

Impedance of solenoid

$Z = \dfrac{V_{ac}}{I_{ac}} = \dfrac{100}{0.5} = 200 \ \Omega$

Inductive reactance

$X_L = \sqrt{Z^2 - R^2} = \sqrt{200^2 - 10^2} = 199.75\Omega$

Inductance of solenoid

$L = \dfrac{X_L}{2\pi \ f} = \dfrac{199.75}{2\pi\times 50} = 0.64 \ H$

$0.21 - H$ inductor and a

$88- \Omega$ resistor are connected in series to a

$220-V , 50-Hz$ AC source. The current in the circuit and the phase angle between the current and the source voltage are respectively. Use

$\pi = 22/7$

0%
$2 A, tan^{-1} 3/4$
0%
$14.4 A, tan^{-1} 7/8$
0%
$14.4 A, tan^{-1} 8/7$
0%
$3.28 A, tan^{-1} 2/11$

Explanation

The impedance of the circuit is given as,

$Z = \sqrt {{R^2} + {X_L}^2}$

$Z = \sqrt {{{\left( {88} \right)}^2} + {{\left( {2 \times \pi \times 50 \times 0.21} \right)}^2}}$

$Z = 110\;\Omega$

The current in the circuit is given as,

$I = \dfrac{V}{Z}$

$I = \dfrac{{220}}{{110}}$

$= 2\;{\rm{A}}$

The phase angle is given as,

${\rm{\theta }} = {\tan ^{ - 1}}\left( {\dfrac{{{X_L}}}{R}} \right)$

${\rm{\theta }} = {\tan ^{ - 1}}\left( {\dfrac{{2 \times \pi \times 50 \times 0.21}}{{88}}} \right)$

$= {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$

Thus, the current in the circuit is $2\;{\rm{A}}$ and the phase angle is ${\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$ .

In a series L-C-R circuit the voltage across the resistance , capacitance and inductance is 10 V each. If capacitance is short circuited, the voltage across the inductance will be:

0%
$10\,V$
0%
$10\sqrt {3}V$
0%
$10/\sqrt {2}V$
0%
$20\,V$

Explanation

The voltages across all the three components are equal, so the impedance will be the same.

$R = {X_L}$

The current is given as,

$I = \dfrac{V}{{\sqrt {{R^2} + {{\left( {{X_L}} \right)}^2}} }}$

$I = \dfrac{{10}}{{R\sqrt 2 }}$

The potential drop across inductor is given as,

${V_L} = I{X_L}$

${V_L} = IR$

${V_L} = \dfrac{{10}}{{\sqrt 2 }}\;{\rm{V}}$

In the circuit shown in figure when the frequency of oscillator in increase, the reading of ammeter

${A}_{4}$ is same as that of ammeter:

0%
${A}_{1}$
0%
${A}_{2}$
0%
${A}_{3}$
0%
${A}_{1}$ and ${A}_{2}$

An inductor of reactance

$X_L=4\Omega$ and resistor of resistance

$R=3\Omega$ are connected in series with a voltage source of emf

$\varepsilon =(20V)[\sin (100\pi rad/s)t]$ . The current in the circuit at any time t will be?

0%
$I=(4A)[\sin (100\pi rad/s)t+37^o]$
0%
$I=(4A)[\sin (100\pi rad/s)t-37^o]$
0%
$I=(4A)[\sin (100\pi rad/s)t+53^o]$
0%
$I=(4A)[\sin (100\pi rad/s)t-53^o]$

Explanation

The inductor and the resistor are in series. Given,

$R_{ L }=4\Omega ,\quad R=3\Omega$

Resultant impedance R=

$\sqrt { { R }_{ L }^{ 2 }+{ R }^{ 2 } } \\ =\sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } \\ \quad =5\Omega$

In RL circuit, current due to inductor leads by resistance.

Therefore phase angle,

$\phi ={ tan }^{ -1 }\dfrac { 3 }{ 4 }$

${ 37 }^{ 0 }$

Peak voltage,

${ e }_{ o }=20V$

Current, I=

$\dfrac { { e }_{ 0 } }{ R } =\dfrac { 20 }{ 5 } \\ =4A$

Since the current is leading. Therefore value of I is

$4A[sin(100\pi )t+{ 37 }^{ 0 }]$

1144587_1098510_ans_be55056273f640c58f51238e38eb507c.png

$LCR$ oscillation circuit resistance is

$10\Omega$ and inductive reactance at resonace condition is

$1\ k\Omega$ . After how many oscillation peak value os current will fall to

$(\dfrac {1}{e})$ times maximum value of peak current.

0%
$\dfrac {50}{\pi}$
0%
$100$
0%
$50$
0%
$\dfrac {100}{\pi}$

A coil of

$L=5x10^{-3}H$ and

$R=18\ \Omega$ is abruptly supplied a potential of

$5$ volts .What will be the rate of change of current in

$0.001$ second?

$(e^{-3.6}-0.0273)$

0%
$27.3\ amp/sec$
0%
$27.8\ amp/sec$
0%
$2.73\ amp/sec$
0%
$2.78\ amp/sec$

In the given circuit find the ratio of

$i_{1}$ to

$i_{2}$ . Where

$i_{1}$ is the initial (at

$t = 0)$ current, and

$i_{2}$ is steady state (at

$t = \infty)$ current through the battery.

0%
$1.0$
0%
$0.8$
0%
$1.2$
0%
$1.5$

Explanation

Inductor is open circuited at

$t=0$ and short circuited at

$t=\infty$

Hence,

${I}_{1}=\dfrac{10V}{10R}=1$ A

${I}_{2}=\dfrac{10V}{8R}=\dfrac{10}{8}$ A

So,

$\dfrac{{I}_{1}}{{I}_{2}}=\dfrac{1}{\dfrac{10}{8}}=\dfrac{8}{10}=0.8$

In a series

$RLC$ circuit, potential difference across

$R,L$ and

$C$ are

$30V,60V$ and

$100V$ respectively as shown in figure. The emf of source (in volts)is:

0%
$190$
0%
$70$
0%
$50$
0%
$40$

A condenser of capacity

$20 \mu F$ is first charged and then discharged through a

$10 mH$ inductance. Neglecting the resistance of the coil, the frequency of the resulting vibrations will be

0%
$356$ cycle/s
0%
$35.6$ cycle/s
0%
$356 \times 10^3$ cycle/s
0%
$3.56$ cycle/s

Explanation

The frequency of the resulting vibrations is given as,

$f = \dfrac{1}{{2\pi \sqrt {LC} }}$

$f = \dfrac{1}{{2\pi \sqrt {10 \times {{10}^{ - 3}} \times 20 \times {{10}^{ - 6}}} }}$

$f = 356\;{\rm{cycle/s}}$

$LCR$ series circuit with

$R=100\Omega$ is connected to a

$200\ V,50\ Hz$ a.c source. When only the capacitance is removed, the current leads the voltage by

$60^{o}$ . When only the inductance is removed, the current leads the voltage by

$60^{o}$ . The current in the circuit is :

0%
$2A$
0%
$1A$
0%
$\dfrac {\sqrt {3}}{2}A$
0%
$\dfrac {2}{\sqrt {3}}A$

Explanation

Given,

$R=100\Omega$

$V=200V$

$f=50Hz$

$\phi=60^0$

When a capacitance is removed, the current leads the voltage by

$\phi$

$tan\phi=\dfrac{X_L}{R}$

$X_L=R tan60=\sqrt{3}R$ . . . .(1)

When the inductance is removed current leads the voltage by

$\phi$

$tan\phi=\dfrac{X_C}{R}$

$X_C=\sqrt{3}R$ . . . . . .(2)

From equation (1) and (2), we get

$X_C=X_L$ (this is the condition of resonance)

And at the resonance,

$Z=R$

The current in the circuit is

$I=\dfrac{V}{R}$

$I=\dfrac{200}{100}$

$I=2A$

The correct option is A.

In an LCR circuit, the resonating frequency is

$500$

$kH_z$ . If the value of

$L$ is doubled and value of

$C$ is decreased to

$\cfrac{1}{8}$ times of its initial values, then the new resonating frequency in

$kH_z$ will be

0%
$250$
0%
$500$
0%
$1000$
0%
$2000$

Explanation

$\textbf{Step 1 - Formula of resonant frequency,}$

$f = \dfrac {1}{2\pi \sqrt {LC}}$

Given,

$L_{2} = 2L_{1}$

$C_{2} = \dfrac {C_{1}}{8}$

$f_{1} = 500\ kH_{2}$

$\textbf{Step 2 - Comparing two resonant frequency,}$

$\dfrac {f_{1}}{f_{2}} = \dfrac {\dfrac {1}{2\pi \sqrt {L_{1}C_{1}}}}{\dfrac {1}{2\pi \sqrt {L_{2}C_{2}}}} = \sqrt {\dfrac {L_{2}C_{2}}{L_{1}C_{1}}}$

$\dfrac {f_{1}}{f_{2}} = \sqrt {\dfrac {1}{4}} = \dfrac {1}{2}$

$f_{2} = 2f_{1} = 2\times 500\ kH_{2}$

$f_{2} = 1000\ kH_{2}$

A coil has resistance

$30 \ ohm$ and inductive reactance

$20 \ ohm$ at 50 Hz frequency. If an ac source of 200 V,100 Hz is connected across the coil, the current in the coil will be

0%
$20\sqrt 13 A$
0%
2.0 A
0%
4.0 A
0%
none

In L-C oscillation if frequency of oscillation of charge is

$f$ , then frequency of oscillation of magnetic energy is

0%
$f$
0%
$2f$
0%
$\cfrac{f}{2}$
0%
$4f$

An inductor of inductance

$2.0$

$H$ and a resistor of resistance

$10$

$\Omega$ are connected senes to a battery of EMF

$20$

$V$ in a circuit as shown.The key

${ K }_{ 1 }$ been kept closed for a long time. Then at

$t$ =

$0$ ,

${ K }_{ 1 }$ i s opened and key

${ K }_{ 2 }$ is closed simultaneously, The rate of decrease of current in the circuit at

$t$ =

$1.0 s$ wili be (

${ e }^{ 5 }$ =

$150$ )

0%
$\dfrac { 1 }{ 15 }$ $A/s$
0%
$\dfrac { 2 }{ 15 }$ $A/s$
0%
$\dfrac { 1 }{ 5 }$ $A/s$
0%
$\dfrac { 4 }{ 15 }$ $A/s$

A capacitor of capacitance

$C$ has initial charge

${ Q }_{ 0 }$ and connected to an inductor of inductance

$L$ as shown. At

$t=0$ switch

$S$ is closed. The current through the inductor when energy in the capacitor is three tirnes the energy of inductor is

0%
$\dfrac { { Q }_{ 0 } }{ 2\sqrt { LC } }$
0%
$\dfrac { { Q }_{ 0 } }{ \sqrt { LC } }$
0%
$\dfrac { 2{ Q }_{ 0 } }{ \sqrt { LC } }$
0%
$\dfrac { 4{ Q }_{ 0 } }{ \sqrt { LC } }$

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 13 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Alternating Current Quiz 13 - MCQExams.com

0:0:1

Practice Class 12 Medical Physics Quiz Questions and Answers

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