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 A. When another alternating current passes through the circuit, the AC ammeter reads 8 A. Then the reading of this ammeter if DC and AC flow through the circuit simultaneously is:

1. $10\sqrt{2}$ A

2. 14 A

3. 10 A

4. 15 A

If q is the capacitor's charge and i is the current at time t, the voltage V will be:

A direct current of 5 A is superimposed on an alternating current I = 10$\sin \omega t$ flowing through a wire. The effective value of the resulting current will be:

1. \(15/2~A\)

2. $5\sqrt{\mathrm{3 }}A$

3. $5\sqrt{\mathrm{5 }}A$

4. 15 A

An L-C-R series circuit with 100 Ω resistance is connected to an AC source of 200 V and an angular frequency of 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by $60°$. When only the inductance is removed, the current leads the voltage by $60°$. Calculate the power dissipated in the L-C-R circuit.

1. 200 W

2. 400 W

3. 300 W

4. Zero

An ideal resistance R, ideal inductance L, ideal capacitance C, and AC voltmeters ${\mathrm{V}}_1,$ ${\mathrm{V}}_2,$ ${\mathrm{V}}_3$ $\mathrm{and}$ ${\mathrm{V}}_4$ are connected to an AC source as shown. At resonance:

A transistor-oscillator using a resonant circuit with an inductance L (of negligible resistance) and a capacitance C has a frequency f. If L is doubled and C is changed to 4C, the frequency will be:

1. f/4

2. 8f

3. f/(2√2)

4. f/2

In an L-C-R series AC circuit, the voltage across each of the components - L, C and R is 50 V. The voltage across the L-R combination will be:
1. 50 V
2. \(50 \sqrt{2} ~V\)
3. 100 V
4. 0 V 

The AC source in the circuit shown in the figure produces a voltage V = 20cos(2000t) volts. Neglecting source resistance, the voltmeter and ammeter readings will be (approximately):

1. 4 V, 2.0 A

2. 0 V, 2 A

3. 0 V, 1.4 A

4. 8 V, 2.0 A

In the circuit shown, the AC source has a voltage
V = 20 cos(ωt) volts with ω= 2000 rad/s. The amplitude of the current will be nearest to:

1. 2 A

2. 3.3 A

3. 2/$\sqrt{5}$ A

4. $\sqrt{5}$ A

In a heating arrangement, an alternating current having a peak value of \(28~\mathrm{A}\) is used. To produce the same heat energy, if direct current is used to produce the same amount of heat, then its magnitude will be:
1. about \(14~\mathrm{A}\) 
2. about \(28~\mathrm{A}\) 
3. about \(20~\mathrm{A}\) 
4. cannot say