Alternating Current Test

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Alternating Current Test - 76

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Question 1
1 / -0

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

A
$$0.05\ s$$

B
$$0.1\ s$$

C
$$0.15\ s$$

D
$$0.3\ s$$

Solution

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.
Question 2
1 / -0

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

A
The sum of the maximum voltages across the elements.

B
The voltage drop across the resistor.

C
The sum of the instantaneous voltages across the elements.

D
The sum of the rms voltages across the elements.

Solution

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

$${V_s} = \sqrt {V_R^2 + {{\left( {{V_L} - {V_c}} \right)}^2}} $$
Question 3
1 / -0

In the given figure, which voltmeter will read zero voltage at resonant frequency $$\omega$$ rad/sec?

A
$${V}_{1}$$

B
$${V}_{2}$$

C
$${V}_{3}$$

D
$${V}_{4}$$

Solution

At Resonant frequency,
$$X_L = X_C$$
multiply both sides by $$I_{rms}$$,
$$V_{L(rms)}=V_{C(rms)}$$ (numerically)
Since we know,
Voltmeter reads rms voltage.
Potential difference across $$V_4=\sqrt{V_{L(rms)}^2-V_{C(rms)}^2}$$ (as $$V_L$$ and $$V_C$$ are $$\pi$$ radian out of phase).
Hence, Potential difference across $$V_4 = 0$$

Option $$D$$ is the correct answer.
Question 4
1 / -0

A $$100\ volt$$ $$AC$$ 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

A
$$\dfrac {100}{\sqrt {2}}volt$$

B
$$100\ volt$$

C
$$50\ volt$$

D
$$50\sqrt {3} $$

Solution

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 $$100\;{\rm{V}}$$.
Question 5
1 / -0

The instantaneous voltages at three terminals marked X,Y and Z are given by
$$V_x=V_0\sin { \omega t } ,$$
$$V_x=V_0\sin { \sin { \left( \omega t+\dfrac { 2\pi }{ 3 } \right) } } $$ and
$$V_x=V_0\sin { \sin { \left( \omega t+\dfrac { 4\pi }{ 3 } \right) } } $$
An ideal voltmeter is configured to read rms value of the potential difference between its terminal. It is connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be

A
$${ V }_{ YZ }^{ rms }={ V }_{ o }\sqrt { \dfrac { 1 }{ 2 } } $$

B
$${ V }_{ XY }^{ rms }={ V }_{ o }\sqrt { \dfrac { 3 }{ 2 } } $$

C
Independent of the choice of the two terminals

D
$${ V }_{ XY }^{ rms }={ V }_{ o }$$

Solution

$$V_{XY}=V_X-V_Y$$
$$V_{XY}={ \left( { V }_{ XY } \right) }_{ 0 }\sin { \left( \omega t+{ Q }_{ 1 } \right) } $$
Where $${ \left( { V }_{ XY } \right) }_{ 0 }=\sqrt { { V }_{ 0 }^{ 2 }+{ V }_{ 0 }^{ 2 }-{ 2V }_{ 0 }^{ 2 }\cos { \dfrac { 2\pi }{ 3 } } } =\sqrt { 3 } { V }_{ 0 }$$
and $${ V }_{ XY }^{ rms }=\dfrac { { \left( { V }_{ XY } \right) }_{ 0 } }{ \sqrt { 2 } } =\sqrt { \dfrac { 3 }{ 2 } } { V }_{ 0 }$$
$$V_{YZ}=V_Y-V_Z$$
$$V_{YZ}={ \left( { V }_{ YZ } \right) }_{ 0 }\sin { \left( \omega t+{ Q }_{ 2 } \right) } $$
Where $${ \left( { V }_{ YZ } \right) }_{ 0 }=\sqrt { { V }_{ 0 }^{ 2 }+{ V }_{ 0 }^{ 2 }-{ 2V }_{ 0 }^{ 2 }\cos { \dfrac { 2\pi }{ 3 } } } =\sqrt { 3 } { V }_{ 0 }$$
and $${ V }_{ YZ }^{ rms }=\dfrac { { \left( { V }_{ YZ } \right) }_{ 0 } }{ \sqrt { 2 } } =\sqrt { \dfrac { 3 }{ 2 } } { V }_{ 0 }$$
Question 6
1 / -0

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.

A
$$200\Omega, 0.64\ H$$

B
$$100\Omega, 0.86\ H$$

C
$$200\Omega, 1.0\ H$$

D
$$100\Omega, 0.93\ H$$

Solution

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   $$L = \dfrac{X_L}{2\pi \ f} = \dfrac{199.75}{2\pi\times 50} = 0.64 \ H$$
Question 7
1 / -0

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:

A
$$10\,V$$

B
$$10\sqrt {3}V$$

C
$$10/\sqrt {2}V$$

D
$$20\,V$$

Solution

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,

$${V_L} = I{X_L}$$

$${V_L} = IR$$

$${V_L} = \dfrac{{10}}{{\sqrt 2 }}\;{\rm{V}}$$
Question 8
1 / -0

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:

A
$${A}_{1}$$

B
$${A}_{2}$$

C
$${A}_{3}$$

D
$${A}_{1}$$ and $${A}_{2}$$

Solution
Question 9
1 / -0

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?

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

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

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

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

Solution

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 $$4A[sin(100\pi )t+{ 37 }^{ 0 }]$$
Question 10
1 / -0

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.

A
$$1.0$$

B
$$0.8$$

C
$$1.2$$

D
$$1.5$$

Solution

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