In an AC circuit, the current is given by; \(i=5\sin\left(100t-\frac{\pi}{2}\right)\) and the AC potential is \(V =200\sin(100 t)~\text V.\) The power consumption is:
1. \(20~\text W\)
2. \(40~\text W\)
3. \(1000~\text W\)
4. zero

Subtopic:  Power factor |
 86%
Level 1: 80%+

To unlock all the explanations of this course, you need to be enrolled.

Hints
Links

To unlock all the explanations of this course, you need to be enrolled.


A coil of inductive reactance of \(31~\Omega\) has a resistance of \(8~\Omega\). It is placed in series with a condenser of capacitive reactance \(25~\Omega\). The combination is connected to an AC source of \(110\) V. The power factor of the circuit is:
1. \(0.56\)
2. \(0.64\)
3. \(0.80\)
4. \(0.33\)

Subtopic:  Power factor |
 86%
Level 1: 80%+
AIPMT - 2006

To unlock all the explanations of this course, you need to be enrolled.

Hints
Links

To unlock all the explanations of this course, you need to be enrolled.


The power factor of the given circuit is:
             

1. \(1 \over 2\) 2. \(1 \over \sqrt2\)
3. \(\sqrt3 \over 2\) 4. \(0\)
Subtopic:  Power factor |
 84%
Level 1: 80%+

To unlock all the explanations of this course, you need to be enrolled.

Hints
Links

To unlock all the explanations of this course, you need to be enrolled.


advertisementadvertisement

An inductor of inductance \(L\) and resistor of resistance \(R\) are joined in series and connected by a source of frequency \(\omega\). The power dissipated in the circuit is:
1. \(\dfrac{\left( R^{2} +\omega^{2} L^{2} \right)}{V}\)

2. \(\dfrac{V^{2} R}{\left(R^{2} + \omega^{2} L^{2} \right)}\)

3. \(\dfrac{V}{\left(R^{2} + \omega^{2} L^{2}\right)}\)

4. \(\dfrac{\sqrt{R^{2} + \omega^{2} L^{2}}}{V^{2}}\)

Subtopic:  Power factor |
 80%
Level 1: 80%+

To unlock all the explanations of this course, you need to be enrolled.

Hints
Links

To unlock all the explanations of this course, you need to be enrolled.


What is the average power dissipated in the AC circuit if current \(i = 100\sin(100t)\) A and \(V = 100\sin\left(100t+\frac{\pi}{3}\right)\) volts?
1. \(2500\) W 2. \(250\) W
3. \(5000\) W 4. \(4000\) W
Subtopic:  Power factor |
 78%
Level 2: 60%+

To unlock all the explanations of this course, you need to be enrolled.

Hints
Links

To unlock all the explanations of this course, you need to be enrolled.


premium feature crown icon
Unlock IMPORTANT QUESTION
This question was bookmarked by 5 NEET 2025 toppers during their NEETprep journey. Get Target Batch to see this question.
✨ Perfect for quick revision & accuracy boost
Buy Target Batch
Access all premium questions instantly
A resistance \(R\) draws power \(P\) when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes \(Z\) the power drawn will be:
\(1 .\) \(P \left(\frac{R}{Z}\right)^{2}\)
\(2 .\) \(P \sqrt{\frac{R}{Z}}\)
\(3 .\) \(P \left(\frac{R}{Z}\right)\)
\(4 .\) \(P\)
Subtopic:  Power factor |
 60%
Level 2: 60%+
NEET - 2015

To unlock all the explanations of this course, you need to be enrolled.

Hints
Links

To unlock all the explanations of this course, you need to be enrolled.


advertisementadvertisement

An AC voltage source is connected to a series \(LCR\) circuit. When \(L\) is removed from the circuit, the phase difference between current and voltage is \(\dfrac{\pi}{3}\). If \(C\) is instead removed from the circuit, the phase difference is again \(\dfrac{\pi}{3}\) between current and voltage. The power factor of the circuit is:
1. \(0.5\)
2. \(1.0\)
3. \(-1.0\)
4. zero

Subtopic:  Power factor |
 67%
Level 2: 60%+
NEET - 2020

To unlock all the explanations of this course, you need to be enrolled.

Hints
Links

To unlock all the explanations of this course, you need to be enrolled.