Alternating Current Online Objective Test | Alternating Current Online Objective Test

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q1. An AC voltage source of variable angular frequency 'w' and fixed amplitude V0 is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When 'w' is increased:

A. The lamp glows dimmer and dimmer

B. The lamp does not glow

C. The lamp starts turning on and off

D. The lamp glows brighter

Answer : Option D
Explaination / Solution:

in CR circuit, impedance

$Z = \sqrt{R^{2} + \frac{1}{ω^{2} C^{2}}}$

current $i = \frac{V}{Z}$

as $ω$ increases, Z decreases. Hence the current i increases. Therefore, the bulb glows brighter.

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q2. 25.0 μF capacitor is connected to a 220 V, 50 Hz source. Capacitive reactance and the current (rms) in the capacitor are circuit

A. 102 Ω , 1.03 A

B. 127 Ω, 1.73 A

C. 111 Ω , 1.73 A

D. 127 Ω , 1.23 A

Answer : Option B
Explaination / Solution:

$\begin{aligned} C = 25 μ F \\ V = 220 v o l t \\ f = 50 H z \end{aligned}$

Capacitive reactance

$\begin{aligned} X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C} = \frac{1}{2 \times 3.14 \times 50 \times 25 \times 10^{- 6}} \\ X_{C} = 127 Ω \end{aligned}$

rms current

$i = \frac{V}{X_{C}} = \frac{220}{127} = 1.73 A$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q3. A light bulb and an open coil inductor are connected to an ac source . If an iron rod is inserted into the interior of the inductor then

A. The bulb glows brighter

B. The bulb glow is unaffected

C. The bulb glows dimmer

D. The bulb glows on and off

Answer : Option C
Explaination / 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 inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q4. A series LCR circuit with R = 20

Ω

, L = 1.5 H and C = 35

μ F

is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit average power transferred to the circuit in one complete cycle is

A. 2,800 W

B. 1,800 W

C. 1,000 W

D. 2,000 W

Answer : Option D
Explaination / Solution:

When the frequency of the supply equals the natural frequency of the circuit

$X_{L} = X_{C}$

in this case impedance Z = R

average power

Pav=Vrms×irms×cosϕ=Vrms×VrmsZ×RZZ=RPav=(Vrms)2R=200×20020=2000W

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q5. Series LCR circuit is connected to a variable frequency 230 V source. L = 5.0 H, C = 80 μF, R = 40 Ω. Impedance of the circuit and the amplitude of current at the resonating frequency are

A. 40 Ω, 8.1 A

B. 40 Ω, 5.75 A

C. 40 Ω, 8.1 A

D. 40 Ω , 8.1 A

Answer : Option B
Explaination / Solution:

$\begin{aligned} R = 40 Ω \\ L = 5 H \\ C = 80 μ F \end{aligned}$

impedence in series LCR circuit $Z = \sqrt{R^{2} + (w L - 1 / w C)^{2}}$

at resonance $ω L = \frac{1}{ω C}$

hence $Z = \sqrt{R^{2} + 0} = R = 40 Ω$

V = 230 Volt

Hence current

$i = \frac{V}{Z} = \frac{230}{40} = 5.75 A$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q6. A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible. Average powers transferred to the inductor and to the capacitor are respectively

A. 10 W, 10 W

B. 10 W, 10 W

C. 0 W,0 W

D. 20 W, 10 W

Answer : Option C
Explaination / Solution:

$P = V I \cos ϕ$

average power consumed by inductor is zero as actual voltage leads the current by $\frac{π}{2}$

average power consumed by capacitor is zero as actual voltage lags the current by $\frac{π}{2}$

$(\cos \frac{π}{2} = 0)$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q7. Angular resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 μF, and R = 7.4 Ω are respectively

A. 111 rad/s 35

B. 90 rad/s 45 Hz

C. 100 rad/s 45

D. 111 rad/s 45

Answer : Option D
Explaination / Solution:

L = 3.0 H

C = 27 $μ F$

R = 7.4 $Ω$

Angular resonant frequency

$ω = \frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{3 \times 27 \times 10^{- 6}}} = \frac{1}{9 \times 10^{- 3}} = 111 r a d / s$

Q-factor

$Q = \frac{1}{R} \sqrt{\frac{L}{C}} = \frac{1}{7.4} \sqrt{\frac{3}{27 \times 10^{- 6}}} = 45$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q8. A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. Number of turns in the secondary in order to get output power at 230 V is

A. 325

B. 400

C. 425

D. 380

Answer : Option B
Explaination / Solution:

Np = no. of turns in primary coil = 4000

Ns = no. of turns in primary coil

Vp = input voltage = 2300 V

Vs = output voltage = 230 V

$\begin{aligned} \frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} \\ \frac{230}{2300} = \frac{N_{s}}{4000} \\ N_{s} = 400 \end{aligned}$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q9. A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 ΩΩ per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town. Line power loss in the form of heat is

A. 500 kW

B. 600 kW

C. 550 kW

D. 450 kW

Answer : Option B
Explaination / Solution:

Total electric power required, P = 800 kW

supply voltage V = 220 volt

voltage at which electric plant generating power V1 = 440 volt

distance between town and power generating station, d = 15 km

total resistance of two wire cable R = (15+15) x 0.5 = 15 ohm

input voltage of transformer Vi = 4000 volt

output voltage of transformer Vo = 220 volt

current in wire line

$i = \frac{P}{V_{i}} = \frac{800 \times 10^{3}}{4000} = 200 A$

line power loss

$P_{l o s s} = i^{2} R = 200 \times 200 \times 15 = 600 \times 10^{3} = 600 K W$

# Alternating Current # CBSE 12TH PHYSICS Prepare / Learn

Q10. A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5 Ω per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town. Power that the plant supply, assuming there is negligible power loss due to leakage is

A. 1400 kW

B. 1000 kW

C. 1300 kW

D. 1200 kW

Answer : Option A
Explaination / Solution:

Total electric power required, P = 800 kW

supply voltage V = 220 volt

voltage at which electric plant generating power V1 = 440 volt

distance between town and power generating station, d = 15 km

total resistance of two wire cable R = (15+15) x 0.5 = 15 ohm

input voltage of transformer Vi = 4000 volt

output voltage of transformer Vo = 220 volt

current in wire line

$i = \frac{P}{V_{i}} = \frac{800 \times 10^{3}}{4000} = 200 A$

$P_{l o s s} = i^{2} R = 200 \times 200 \times 15 = 600 \times 10^{3} = 600 K W$

total power supplied by the plant = 800 + 600 = 1400 KW

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