Alternating Current Test

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Alternating Current Test - 18

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Question 1
1 / -0

An inductor $20$ mH, a capacitor $100$ $\mu$ F and a resistor $50$ $\Omega$ are connected in series across a source of emf, V $=10$ $\sin 314$ t. The power loss in the circuit is?

A
$2.74$ W

B
$0.79$ W

C
$1.13$ W

D
$0.43$ W

Solution

$L = 20 mH$ $C = 100 \mu F$ $R = 50 \Omega$
$V = 10 sin(314 t)$
$V_0= 10$ , $\omega = 314$
$X_L = wL= 314 \times 20\times 10^{-3}= 6.28 \Omega$
$X_C = \dfrac{1}{\omega C}=31.8 \Omega$
$Z = \sqrt{R^2 + (X_C- X_L)^2} = 56.1$
$Power \ loss P=\dfrac{V_0^2 R}{2 Z^2}= 0.79 W$
Question 2
1 / -0

A coil has resistance $30 ohm$ and inductive reactance $20 ohm$ at $50 Hz$ frequency. If an $ac$ source of $200 volt, 100 Hz$ , is connected across the coil, the current in the coil will be

A
$4.0 A$

B
$8.0 A$

C
$\displaystyle\frac{20}{\sqrt{13}} A$

D
$2.0 A$

Solution

If $\omega = 50 \times 2\pi$ then $\omega L = 20\Omega$
If ${\omega}^{\prime} = 100 \times 2\pi$ then ${\omega}^{\prime} L = 40\Omega$
$I = \displaystyle\frac{200}{Z} \displaystyle\frac{200}{\sqrt{{R}^{2} + {\left({\omega}^{\prime}L\right)}^{2}}} = \displaystyle\frac{200}{\sqrt{{30}^{2} + {\left(40\right)}^{2}}}$
$I = 4 A$ .
Question 3
1 / -0

A coil of self-inductance L is connected in series with a bulb B and an AC source. Brightness of the bulb decreases when

A
Frequency of the AC source is decreased

B
Number of turns in the coil is reduced

C
A capacitance of reactance $X_{C}=X_{L}$ is included in the same circuit

D
An iron rod is inserted in the coil

Solution

As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence. the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied AC voltage appears across the lnductor, leaving less voltage across the bulb. Therefore' the brightness of the light bulb decreases.
Question 4
1 / -0

A series R-C circuit is connected to an alternating voltage source. Consider two situation:
(a) When capacitor is filled
(b) When capacitor is mica filled
Current through resister is i and voltage across capacitor is V then :

A
$V_a=V_b$

B
$V_a < V_b$

C
$V_a>V_b$

D
$i_a = i_b$

Solution

For series C-R circuit, the impedance $Z=\sqrt{R^2+X_C^2}$ where $X_C=\frac{j}{\omega C}$ and current $I=V/Z$
When the capacitor is filled by mica, the capacitance will be increased. If C increases, $X_C$ decreases, so the current will increase and
hence voltage across resistance increases and voltage across capacitor decreases. thus, $V_a>V_b$
Question 5
1 / -0

A $40 \mu F$ capacitor is connected to a $200 V,\ 50 Hz \ ac$ supply. The rms value of the current
in the circuit is, nearly :

A
$2.05 A$

B
$2.5 A$

C
$25.1 A$

D
$1.7 A$

Solution

RMS value of applied voltage = $$200\ V$$

Impedance of a capacitor is given by:
$X_c=\dfrac{1}{2\pi f C}$

Hence, rms current through it is:
$I_{rms}=\dfrac{V}{X_C}$

$I_{rms}=200 \times 2 \times \pi \times 50 \times 40 \times 10^{-6}$

$I_{rms}=2.51 \ A$
Question 6
1 / -0

A condenser of $250\mu F$ is connected in parallel to a cell of inductance $0.16$ mH, while its effective resistance is while its. Determine the resonant frequency.

A
$9\times 10^4$ Hz

B
$16\times 10^7$ Hz

C
$8\times 10^5$ Hz

D
$9\times 10^3$ Hz

Solution

Given
Capacitance, $C= 250\mu F= 250\times 10^{-6} F$
Inductance , $L=0.16mH= 0.16 \times 10^{-3}H$

Resonant frequency , $f=\dfrac1{2\pi \sqrt{LC}}= \dfrac1{2\times 3.14\times \sqrt{250\times 10^{-6} \times 0.16 \times 10^{-3}}}= 8\times 10^5 Hz$
Question 7
1 / -0

In a series LCR circuit, the voltage across the resistance, capacitance and inductance is $10\ V$ each. If the capacitance is short circuited the voltage across the inductance will be :

A
$10\ V$

B
$10\sqrt{2}\ V$

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

D
$20\ V$

Solution

Since the voltages are all equal, the impedance of all the elements is same.

$R=X_L=X_C$

Total voltage in the circuit= $I\sqrt{R^2+(X_L-X_C)^2}=IR=10V$

When capacitance is shorted,

$I=\dfrac{V}{\sqrt{R^2+X_L^2}}=\dfrac{10}{\sqrt{2}R}$

Potential drop across inductor= $IX_L=IR=\dfrac{10}{\sqrt{2}}V$
Question 8
1 / -0

Reciprocal of Impedance is:

A
Susceptance

B
Conductance

C
Admittance

D
Transconductance

Solution

Impedance is the opposition a circuit presents to a current when a voltage is applied.
Admittance is a measure of how easily a circuit or device will allow a current to flow.
Admittance is defined as $Y=\dfrac{1}{Z}$
where $Z$ is the impedance of the circuit.
Question 9
1 / -0

A $10\ \mu F$ capacitor is connected across a $$200\ V$$, $50 \ Hz$ A.C. supply. The peak current through the circuit is :

A
$$0.6\ A$$

B
$0.6\sqrt{2}A$

C
$(0.06\sqrt{2})A$

D
$0.6{\pi}\ A$

Solution

RMS value of applied voltage = $$200\ V$$

Impedance of a capacitor is given by:
$X_C=\dfrac{1}{2\pi f C}$

Hence, rms current through it is:
$I=\dfrac{V}{X_C}$
$I=200 \times 2 \times \pi \times 50 \times 10 \times 10^{-6}$
$I=0.2 \pi \ A$

Peak current, $I_{max}=\sqrt 2 I$
$=0.628 \sqrt 2\ A$
Question 10
1 / -0

The instantaneous value of emf and current in an A.C. circuit are $E=1.414$ sin( $100\pi t-\frac{\pi }{4}$ ) , $I=0.707$ sin( $100\pi t$ ) . The resistance of the circuit is

A
$2\Omega$

B
$\sqrt{2}\Omega$

C
$\frac{1}{\sqrt{2}}\Omega$

D
$\frac{1}{2}\Omega$

Solution

resistance = impedance $\times cos(\frac{\pi}{4})$
$= \frac{1.414}{0.707}\times \frac{1}{\sqrt{2}}$
$= \sqrt{2} \Omega$

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Alternating Current Test - 18

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Alternating Current Test - 18

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Question 1

2.742.742.74 W

0.790.790.79 W

1.131.131.13 W

0.430.430.43 W

Solution

Question 2

4.0A4.0 A4.0A

8.0A8.0 A8.0A

2013A\displaystyle\frac{20}{\sqrt{13}} A13​20​A

2.0A2.0 A2.0A

Solution

Question 3

Frequency of the AC source is decreased

Number of turns in the coil is reduced

A capacitance of reactance XC=XLX_{C}=X_{L}XC​=XL​ is included in the same circuit

An iron rod is inserted in the coil

Solution

Question 4

Va=VbV_a=V_bVa​=Vb​

Va<VbV_a < V_bVa​<Vb​

Va>VbV_a>V_bVa​>Vb​

ia=ibi_a = i_bia​=ib​

Solution

Question 5

2.05A2.05 A2.05A

2.5A2.5 A2.5A

25.1A25.1 A25.1A

1.7A1.7 A1.7A

Solution

Question 6

9×1049\times 10^49×104Hz

16×10716\times 10^716×107 Hz

8×1058\times 10^58×105 Hz

9×1039\times 10^39×103 Hz

Solution

Question 7

10 V10\ V10 V

102 V10\sqrt{2}\ V102​ V

102 V\dfrac{10}{\sqrt{2}} \ V2​10​ V

20 V20\ V20 V

Solution

Question 8

Susceptance

Conductance

Admittance

Transconductance

Solution

Question 9

$$0.6\ A$$

0.62A0.6\sqrt{2}A0.62​A

(0.062)A(0.06\sqrt{2})A(0.062​)A

0.6π A0.6{\pi}\ A0.6π A

Solution

Question 10

2Ω2\Omega 2Ω

2Ω\sqrt{2}\Omega 2​Ω

12Ω\frac{1}{\sqrt{2}}\Omega2​1​Ω

12Ω\frac{1}{2}\Omega21​Ω

Solution

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$2.74$ W

$0.79$ W

$1.13$ W

$0.43$ W

$4.0 A$

$8.0 A$

$\displaystyle\frac{20}{\sqrt{13}} A$

$2.0 A$

A capacitance of reactance $X_{C}=X_{L}$ is included in the same circuit

$V_a=V_b$

$V_a < V_b$

$V_a>V_b$

$i_a = i_b$

$2.05 A$

$2.5 A$

$25.1 A$

$1.7 A$

$9\times 10^4$ Hz

$16\times 10^7$ Hz

$8\times 10^5$ Hz

$9\times 10^3$ Hz

$10\ V$

$10\sqrt{2}\ V$

$\dfrac{10}{\sqrt{2}} \ V$

$20\ V$

$0.6\sqrt{2}A$

$(0.06\sqrt{2})A$

$0.6{\pi}\ A$

$2\Omega$

$\sqrt{2}\Omega$

$\frac{1}{\sqrt{2}}\Omega$

$\frac{1}{2}\Omega$