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Given that the current \(i_1=3A \sin \omega t\) and the current \(i_2=4A \cos \omega t,\) what will be the expression for the current \(i_3\)?

                  
1. \(5 A \sin \left(\omega t+53^{\circ}\right) \)
2. \(5 A \sin \left(\omega t+37^{\circ}\right) \)
3. \(5 A \sin \left(\omega t+45^{\circ}\right) \)
4. \( 5 A \sin \left(\omega t+30^{\circ}\right)\)

Subtopic:  AC vs DC |
 59%
Level 3: 35%-60%
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In a box \(Z\) of unknown elements (\(L\) or \(R\) or any other combination), an ac voltage \(E = E_0 \sin(\omega t + \phi)\) is applied and the current in the circuit is found to be \(I = I_0 \sin\left(\omega t + \phi +\frac{\pi}{4}\right)\). The unknown elements in the box could be:
             

1. Only the capacitor
2. Inductor and resistor both
3. Either capacitor, resistor, and an inductor or only capacitor and resistor
4. Only the resistor
Subtopic:  Different Types of AC Circuits |
 66%
Level 2: 60%+
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Given below are two statements.
Assertion (A): When a current \(I=(3+4 \sin \omega t)\) flows in a wire, then the reading of
a dc ammeter connected in series is \(4\) units.
Reason (R): A dc ammeter measures only the value of the current amplitude.
 
1. Both (A) and (R) are True and (R) is the correct explanation of (A).
2. Both (A) and (R) are True but (R) is not the correct explanation of (A).
3. (A) is True but (R) is False.
4. Both (A) and (R) are False.
Subtopic:  AC vs DC |
 61%
Level 2: 60%+
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In the transformer shown in the figure, the ratio of the number of turns of the primary to the secondary is \(\dfrac{N_1}{N_2}= \dfrac{1}{50}.\) If a voltage source of \(10~\text V\) is connected across the primary, then the induced current through the load of \(10~\text{k}\Omega\) in the secondary is:
             
1. \(\dfrac{1}{20}~\text{A}\)
2. zero
3. \(\dfrac{1}{10}~\text{A}\)
4. \(\dfrac{1}{5}~\text{A}\)
Subtopic:  Transformer |
 75%
Level 2: 60%+
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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%+
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An alternating current is given as \(i = i_1\cos\omega t- i_2 \sin \omega t.\) The value of rms current is given by:
1. \( \frac{1}{\sqrt{2}}\left(i_1+i_2\right) \)
2. \( \frac{1}{\sqrt{2}}\left(i_i+i_2\right)^2 \)
3. \( \frac{1}{\sqrt{2}}\left(i_1^2+i_2^2\right)^{1 / 2} \)
4. \( \frac{1}{2}\left(i_1^2+i_2^2\right)^{1 / 2}\)
Subtopic:  RMS & Average Values |
 81%
Level 1: 80%+
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An AC ammeter is used to measure the current in a circuit. When a given direct current passes through the circuit, the AC ammeter reads \(6~\text A.\) When another alternating current passes through the circuit, the AC ammeter reads \(8~\text A.\) Then the reading of this ammeter if DC and AC flow through the circuit simultaneously is:
1. \(10 \sqrt{2}~\text A\) 
2. \(14~\text A\) 
3. \(10~\text A\) 
4. \(15~\text A\) 

Subtopic:  AC vs DC |
 67%
Level 2: 60%+
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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
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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
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A direct current of \(5~ A\) is superimposed on an alternating current \(I=10sin ~\omega t\) flowing through a wire. The effective value of the resulting current will be:

1. \(15/2~A\) 2. \(5 \sqrt{3}~A\)
3. \(5 \sqrt{5}~A\) 4. \(15~A\)
Subtopic:  AC vs DC |
 62%
Level 2: 60%+
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