Alternating Current Online Objective Test | Alternating Current Online Objective Test

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q1. The current amplitude in a pure inductor in a radio receiver is to be 250 μA when the voltage amplitude is 3.60 V at a frequency of 1.60 MHz (at the upper end of the AM broadcast band). Inductance needed is

A. 1.13mH

B. 1.43mH

C. 1.23mH

D. 1.33mH

Answer : Option B
Explaination / Solution:

$\begin{aligned} i = 250 μ A = 250 \times 10^{- 6} A \\ V = 3.60 v o l t \end{aligned}$

f = 1.6 MHz = 1.6 x 106 Hz

If only inductor is present in the circuit, then

$V = i X_{L}$

inductive reactance $X_{L} = \frac{V}{i} = \frac{3.60}{250 \times 10^{- 6}} = 14400 Ω$

$X_{L} = ω L = 2 π f L$

inductance

$L = \frac{X_{L}}{2 π f} = \frac{14400}{2 \times 3.14 \times 1.6 \times 10^{6}} = 1.433 \times 10^{- 3} = 1.43 m H$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q2. A 200 ohm resistor is connected in series with a 5μF capacitor. The voltage across the resistor is VR = (1.20 V) cos(2500 rad/s)t . Capacitive reactance is

A. 80 Ω

B. 60 Ω

C. 90 Ω

D. 70 Ω

Answer : Option A
Explaination / Solution:

VR = (1.20 V) cos(2500 rad/s)t

ω=2500rad/s

C=5μF=5×10−6F

Capacitive reactance

XC=1ωC=12500×5×10−6=80Ω

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q3. A hair dryer meant for 110V, 60Hz is to be used in India . If 220 V is the supply voltage in India , the turns ratio for a transformer would be

A. step-down transformer with turn ratio 3:1

B. step-down transformer with turn ratio 2:1

C. step-down transformer with turn ratio 2.5:1

D. step-up transformer with turn ratio 1:2

Answer : Option B
Explaination / Solution:

the ratio between the number of primary turns to the number of secondary turns being called the “turns ratio” or “transformer ratio”.

$\begin{aligned} \frac{N_{p}}{N_{s}} = \frac{V_{p}}{V_{s}} (= \frac{i n p u t}{o u t p u t}) \\ \frac{N_{p}}{N_{s}} = \frac{220}{110} = \frac{2}{1} \end{aligned}$

turn ratio is $N_{p} : N_{s} = 2 : 1$

$N_{p} > N_{s}$ , hence it is a step down transformer.

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q4. Mr Iyer takes a grinder rated 220 V 50 Hz to the U.S where supply is 110 60 Hz . He needs to use a

A. step-up transformer with turn ratio 1:2

B. step-up transformer with turn ratio 1:2.5

C. step-up transformer with turn ratio 1:3.6

D. step-up transformer with turn ratio 1:3

Answer : Option A
Explaination / Solution:

$\begin{aligned} \frac{N_{p}}{N_{s}} = \frac{V_{p}}{V_{s}} (= \frac{i n p u t}{o u t p u t}) \\ \frac{N_{p}}{N_{s}} = \frac{110}{220} = \frac{1}{2} \\ N_{p} : N_{s} = 1 : 2 \end{aligned}$

$N_{p} < N_{s}$ , hence it is a step up transformer.

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q5. You have a special light bulb with a very delicate wire filament. The wire will break if the current in it ever exceeds 1.50 A, even for an instant. What is the largest root-mean-square current you can run through this bulb?

A. 1.46 A

B. 1.26 A

C. 1.56 A

D. 1.06 A

Answer : Option D
Explaination / Solution:

maximum value of current $i_{0} = 1.5 A$

root-mean-square current

$i_{r m s} = \frac{i_{0}}{\sqrt{2}} = \frac{1.50}{\sqrt{2}} = 1.06 A$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q6. The voltage across the terminals of an ac power supply varies with time according to Vcosωt. The voltage amplitude is V = 45.0 V. Root-mean-square potential difference is

A. 37.8 V

B. 35.8 V

C. 33.8 V

D. 31.8 V

Answer : Option D
Explaination / Solution:

maximum voltage across the terminals V = 45 volt

Root-mean-square potential difference

$V_{r m s} = \frac{V}{\sqrt{2}} = \frac{45}{\sqrt{2}} = 31.8 v o l t$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q7. The voltage across the terminals of an ac power supply varies with time according to Vcosωt. The voltage amplitude is V = 45.0 V. Average potential difference between the two terminals of the power supply is

A. 33.8 V

B. zero V

C. 35.8 V

D. 37.8 V

Answer : Option B
Explaination / Solution:

average value of ac voltage for a half cycle is positive and similiarly the mean value of ac voltage for other half cycle is negative.

Average potential difference between the two terminals for complete full cycle

$V_{a v} = (0.637 V_{0}) + (- 0.637 V_{0}) = 0$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q8. An inductor with L = 9.50 mH is connected across an ac source that has voltage amplitude 45.0 V. Phase angle for the source voltage relative to the current is

A. 120

^{\circ}

B. -90

^{\circ}

C. 180

^{\circ}

D. 90

^{\circ}

Answer : Option D
Explaination / Solution:

if only inductor is present in circuit then R = 0

$\tan ϕ = \frac{X_{L}}{R} = \frac{X_{L}}{0} = \infty$

hence, phase angle $ϕ = 90^{\circ}$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q9. An inductor with L = 9.50 mH is connected across an ac source that has voltage amplitude 45.0 V. Frequency of the source that results in a current amplitude of 3.90 A is

A. 150 Hz

B. 193 Hz

C. 129 Hz

D. 180 Hz

Answer : Option B
Explaination / Solution:

$\begin{aligned} V = 45 v o l t \\ L = 9.5 m H \\ i = 3.9 A \\ f = ? \\ V = i X_{L} = i \times ω L = i \times 2 π f L \end{aligned}$

Frequency of the source

$\begin{aligned} f = \frac{V}{i \times 2 π L} = \frac{45}{3.9 \times 2 \times 3.14 \times 9.5 \times 10^{- 3}} = 0.193 \times 10^{3} \\ f = 193 H z \end{aligned}$

$\begin{aligned} V = 45 v o l t \\ L = 9.5 m H \\ i = 3.9 A \\ f = ? \\ V = i X_{L} = i \times ω L = i \times 2 π f L \end{aligned}$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q10. The wiring for a refrigerator contains a starter capacitor. A voltage of amplitude 170 V and frequency 60.0 Hz applied across the capacitor is to produce a current amplitude of 0.850 A through the capacitor. Capacitance required is

A. 23.4 μF

B. 13.3 μF

C. 17.8 μF

D. 15.3 μF

Answer : Option B
Explaination / Solution:

$\begin{aligned} V = 170 v o l t \\ f = 60 H z \\ i = 0.85 A \\ V = i X_{C} = i \frac{1}{ω C} = \frac{i}{2 π f C} \end{aligned}$

Capacitance required

$C = \frac{i}{2 π f V} = \frac{0.85}{2 \times 3.14 \times 60 \times 170} = 13.3 \times 10^{- 6} F = 13.3 μ F$

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