Power in AC Circuits - NEET Physics Chapterwise MCQs & PYQs

NEET Power in AC Circuits MCQs & PYQs

Question 21:

easy

A coil of inductive reactance $31 \Omega$ has a resistance of $8 \Omega$. It is placed in series with a condenser of $25 \Omega$. The combination is connected to an AC source of $110 V$. The power factor of the circuit is:

(2006)

Net reactance is $X = X_L - X_C = 31 - 25 = 6 \Omega$.
Impedance $Z = \sqrt{R^2 + X^2} = \sqrt{8^2 + 6^2} = 10 \Omega$.
Power factor is $\cos\phi = \frac{R}{Z}$.
$\cos\phi = \frac{8}{10} = 0.80$.

Question 22:

easy

For a series LCR circuit the power loss at resonance is:

(2002)

At resonance in a series LCR circuit, $X_L = X_C$, meaning the circuit is purely resistive.
Total impedance $Z = R$.
Power loss is $P = I_{rms}^2 Z \cos\phi$.
Since $\cos\phi = 1$ and $Z = R$, $P = I^2 R$.

Question 23:

easy

In an a.c. circuit with phase voltage V and current I, the power dissipated is

(1997)

The average power dissipated in an A.C. circuit is $P = V_{rms} I_{rms} \cos\phi$.
Here, $\phi$ is the phase angle between the voltage and the current.
The term $\cos\phi$ is known as the power factor.
Therefore, power dissipated depends on the phase angle between V and I.

Question 24:

easy

In an A.C. circuit, the current flowing is $I = 5 \sin (100t – pi/2)$ ampere and the potential difference is $V = 200 \sin (100t)$ volts. The power consumption is equal to

(1995)

The phase difference between voltage $V = 200 \sin(100t)$ and current $I = 5 \sin(100t - \pi/2)$ is $\phi = \pi/2$.
Power factor of the circuit is $\cos\phi = \cos(\pi/2) = 0$.
Average power consumption is $P = V_{rms} I_{rms} \cos\phi$.
$P = V_{rms} I_{rms} \times 0 = 0 W$.

Question 25:

easy

Two cables of copper are of equal lengths. One of them has a single wire of area of cross-section A, while other has 10 wires of cross-sectional area A/10 each. Give their suitability for transporting A.C. and D.C

(1994)

Alternating current (A.C.) exhibits the skin effect, concentrating on the conductor's surface.
Multiple stranded wires have a larger effective surface area, reducing resistance for A.C.
Direct current (D.C.) flows uniformly across the entire cross-section.
Thus, multiple strands are suitable for A.C., and either can be used for D.C.

Question 26:

easy

A series LCR circuit containing $5.0 H$ inductor, $80 \mu F$ capacitor and $40 \Omega$ resistor is connected to 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be:

(2021)

Resonant frequency $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5 \times 80 \times 10^{-6}}} = 50 rad/s$.
Half-power frequencies are $\omega = \omega_0 \pm \Delta\omega$, where $\Delta\omega = \frac{R}{2L}$.
$\Delta\omega = \frac{40}{2 \times 5} = 4 rad/s$.
Therefore, $\omega = 50 \pm 4$, which gives $46 rad/s$ and $54 rad/s$.