Free Objective Test 02 Practice Test - 11th and 12th

Question 1

One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has self inductance of value (f = 50 Hz)

A. 0.052 H

B. 2.42 H

C. 16.2 mH

D. 1.62 mH

SOLUTION

Solution : A

Current through the bulb $i = \frac{P}{V} = \frac{60}{10} = 6 A$

$V = \sqrt{V_{R}^{2} + V_{L}^{2}}$
$(100)^{2} = (10)^{2} + V_{L}^{2} \Rightarrow V_{L} = 99.5$ Volt
Also $V_{L} = i X_{L} = i \times (2 π v L)$
$\Rightarrow 99.5 = 6 \times 2 \times 3.14 \times 50 \times L \Rightarrow L = 0.052$ H

Question 2

In the circuit given below, what will be the reading of the voltmeter

A. 300 V

B. 900 V

C. 200 V

D. 400 V

SOLUTION

Solution : C

$V^{2} = V_{R}^{2} + (V_{L} - V c)^{2}$

Since $V_{L} = V_{c}$ hence $V = V_{R} = 200$ V

Question 3

In the circuit shown below, what will be the readings of the voltmeter and ammeter

A. 800 V, 2A

B. 220 V, 1.1A

C. 220 V, 2.2 A

D. 100 V, 2A

SOLUTION

Solution : B

$V = V_{R}^{2} + (V_{L} - V_{c})^{2} \Rightarrow V_{R} = V = 220 V$

Also $X_{L} = X_{c},$ Since $(V_{L} - V_{c})$ and current is same

Also $i = \frac{220}{200} = 1.1 A$

Question 4

A bulb and a capacitor are connected in series to a source of alternating current. If its frequency is increased, while keeping the voltage of the source constant, then

A. Bulb will give less intense light

B. Bulb will give more intense light

C. Bulb will give light of same intensity as before

D. Bulb will stop radiating light

SOLUTION

Solution : B

When a bulb and a capacitor are connected in series to an ac source, then on increasing the frequency the current in the circuit is increased, because the impedance of the circuit is decreased. So the bulb will give more intense light.

Question 5

An alternating e.m.f. of angular frequency is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency

\frac{ω}{4}

\frac{ω}{2}

ω

D. 2

ω

SOLUTION

Solution : D

The instantaneous values of emf and current in inductive circuit are given by E= $E_{0}$ sin $ω$ t and $i = i_{0}$ sin $(ω t - \frac{π}{2})$ respectively.

So, $P_{i n s t} = E_{i} = E_{0} s i n ω t \times i_{0} s i n (ω t - \frac{π}{2})$
$= E_{0} i_{0} s i n ω t (s i n ω t c o s \frac{π}{2} - c o s ω t s i n \frac{π}{2})$
$= E_{0} i_{0} s i n ω t c o s ω t$
$= \frac{1}{2} E_{0} i_{0} s i n 2 ω t$ $(s i n 2 ω t = 2 s i n ω t c o s ω t)$
Hence, angular frequency of instantaneous power is 2 $ω$ .

Question 6

The voltage of an ac source varies with time according to the equation V = 200sin $(100 π t) . c o s (100 π t)$ where t is in seconds and V is in volts. Then

A. The peak voltage of the source is 100 volts

B. The peak voltage of the source is 50 volts

C. The peak voltage of the source is

\frac{100}{\sqrt{2}}

volts

D. The frequency of the source is 50 Hz

SOLUTION

Solution : A

$V = 100 \times 2 s i n 100 π t c o s 100 π t = 100 s i n 200 π t$
$\Rightarrow V_{0} = 100$ Volts and Frequency =100Hz

Question 7

In the circuit shown in the figure, the ac source gives a voltage V = 24cos(2000t). Neglecting source resistance, the voltmeter and ammeter reading will be

A. 0V, 0.47A

B. 1.68V, 0.47A

C. 8.46V, 1.4 A

D. 5.6V, 1.4 A

SOLUTION

Solution : C

$Z = \sqrt{(R)^{2} + (X_{L} - X_{c})^{2}}$

$R = 12 Ω, X_{L} = w L = 2000 \times 5 \times 10^{- 3} = 10 Ω$

$X_{c} = \frac{1}{w C} = \frac{1}{2000 \times 50 \times 10^{- 6}} = 10 Ω$ i.e.Z = 10 $Ω$

Maximum current $i_{0} = \frac{V_{0}}{Z} = \frac{24}{12} = 2 A$

Hence $i_{r m s} = \frac{2}{\sqrt{2}} = 1.4 A$

and $V_{r m s} = 6 \times 1.41 = 8.46$ V

Question 8

A telephone wire of length 200 km has a capacitance of 0.014 $μ$ F per km. If it carries an ac of frequency 2.5 kHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum

A. 0.35 mH

B. 35 mH

C. 1.4 mH

D. Zero

SOLUTION

Solution : C

Capacitance of wire
$C = 0.014 \times 10^{- 6} \times 200 = 2.8 \times 10^{- 6} F = 2.8 μ$ F
For impedance of the circuit to be minimum $X_{L} = X_{c} \Rightarrow 2 π v L = \frac{1}{2 π v C}$

$\Rightarrow L = \frac{1}{4 π^{2} v^{2} C} = \frac{1}{4 (3.14)^{2} \times (2.5 \times 10^{3})^{2} \times 2.8 \times 10^{- 6}}$
$= 1.4 \times 10^{- 3}$ H = 1.4mH

Question 9

In a certain circuit current changes with time according to i = 2 $\sqrt{t}$ r.m.s. value of current between t = 2 to t = 4s will be

A. 3A

3 \sqrt{3}

2 \sqrt{3}

(2 - \sqrt{2})

SOLUTION

Solution : C

$^{2} = \frac{\int i^{2} d t}{\int d t} = \frac{\int_{2}^{4} (4 t) d t}{\int_{2}^{4} d t} = \frac{4 \int_{2}^{4} t d t}{2} = 2 {[\frac{t^{2}}{2}]}_{2}^{4} = {[t^{2}]}_{2}^{2} = 12$

$\Rightarrow i_{r m s} = \sqrt{^{2}} = \sqrt{12} = 2 \sqrt{3} A$

Question 10

Match the following

$\begin{matrix} C u r r e n t s & r . m s . v a l u e s (1) x_{0} s i n ω t & (i) x_{0} (2) x_{0} s i n ω t c o s ω t & (i i) \frac{x_{0}}{\sqrt{2}} (3) x_{0} s i n ω t + x_{0} c o s ω t & (i i i) \frac{x_{0}}{(2 \sqrt{2})} \end{matrix}$

A. 1. (i), 2. (ii), 3(i)

B. 1. (ii), 2. (i), 3. (ii)

C. 1. (ii), 2. (iii), 3. (i)

D. 1. (ii), 2. (ii), 3. (i)

SOLUTION

Solution : C

1. rms values $= \frac{x_{0}}{\sqrt{2}}$

2. $x_{0} s i n ω t c o s ω t = \frac{x_{0}}{2} s i n 2 ω t \Rightarrow$ rms value = $\frac{x_{0}}{2 \sqrt{2}}$

3. $x_{0} s i n ω t + x_{0} c o s ω t \Rightarrow$ rms value = $\sqrt{{(\frac{x_{0}}{\sqrt{2}})}^{2} + {(\frac{x_{0}}{\sqrt{2}})}^{2}}$

$= \sqrt{x_{0}^{2}} = x_{0}$

Question 11

The reading of ammeter in the circuit shown will be

A. 2A

B. 2.4 A

C. Zero

D. 1.7 A

SOLUTION

Solution : A

Given $X_{L} = X_{C} = 5 Ω$ this is the condition of resonance. Since $V_{L} = V_{C}$ net voltage across L and C combination will be zero.

Voltage drop across the resistance = 110v. Thus current flowing though Ammeter reading = $\frac{110 v}{55} = 2 A$ .

Question 12

An ac source of angular frequency $ω$ 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 $\frac{ω}{4}$ (but maintaining the same voltage), the current in then circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency $ω$

\sqrt{\frac{3}{5}}

\sqrt{\frac{2}{5}}

\sqrt{\frac{1}{5}}

\frac{1}{2}

SOLUTION

Solution : D

At angular frequency $ω$ , the current in RC circuit is given by

$i_{r m s} = \frac{V_{r m s}}{\sqrt{R^{2} + {(\frac{1}{ω C})}^{2}}} . . . . . . . (i)$

Also $\frac{i_{r m s}}{2} = \frac{V_{r m s}}{\sqrt{R^{2} + {(\frac{1}{\frac{ω}{4} c})}^{2}}} = \frac{V_{r m s}}{\sqrt{R^{2} + \frac{16}{ω^{2} C^{2}}}} . . . . . . (i i)$

From equation (i) and (ii) we get

$3 R^{2} = \frac{12}{ω^{2} C^{2}} \Rightarrow \frac{\frac{1}{ω C}}{R} = \sqrt{\frac{3}{12}} \Rightarrow \frac{X_{c}}{R} = \sqrt{\frac{3}{12}}$

Question 13

An LCR series circuit with a resistance of 100 ohm is connected to an ac source of 100 V (r.m.s.) and angular frequency 300 rad/s. When only the capacitor is removed, the current lags behind the voltage by . When only the inductor is removed the current leads the voltage by . The average power dissipated is

A. 50 W

B. 100 W

C. 200 W

D. 400 W

SOLUTION

Solution : B

$t a n ϕ = \frac{X_{L}}{R} = \frac{X_{C}}{R} \Rightarrow t a n 60^{\circ} = \frac{X_{L}}{R} = \frac{X_{C}}{R}$
$\Rightarrow X_{L} = X_{c} = \sqrt{3} R$
i.e. Z $= \sqrt{R^{2} + (X_{L} - X_{c})^{2}} = R$
So average power $P = \frac{V^{2}}{R} = \frac{100 \times 100}{100} = 100$ W

Question 14

In the adjoining figure the impedance of the circuit will be

A. 120 ohm

B. 50 ohm

C. 60 ohm

D. 45 ohm

SOLUTION

Solution : D

$i_{L} = \frac{90}{30} = 3 A,$ $i_{c} = \frac{90}{90} = 1 A$

Net current through circuit $i = i_{c} - i_{L} = 2 A$

$∴ Z = \frac{V}{i} = \frac{90}{2} = 45 Ω$

Question 15

The voltage of an ac supply varies with time (t) as V = $120 s i n 100 π t c o s 100 π t$ The maximum voltage and frequency respectively are

A. 120 volts, 100 Hz

\frac{120}{\sqrt{2}}

volts , 100 Hz

C. 60 volts, 200 Hz

D. 60 volts, 100 Hz

SOLUTION

Solution : D

$V = 120 s i n 100 π t c o s 100 π t \Rightarrow V = 60 s i n 200 π t$
$V_{m a x} = 60 V$ and Frequency=100Hz

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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