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. 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:
1. | \(\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}+\mathrm{iR}-\frac{\mathrm{q}}{\mathrm{C}}=\mathrm{V}\) |
2. | \(\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}-\mathrm{iR}+\frac{\mathrm{q}}{\mathrm{C}}=\mathrm{V}\) |
3. | \(\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}+\mathrm{iR}+\frac{\mathrm{q}}{\mathrm{C}}=\mathrm{V}\) |
4. | \(\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}-\mathrm{iR}-\frac{\mathrm{q}}{\mathrm{C}}=\mathrm{V}\) |
A direct current of 5 A is superimposed on an alternating current I = 10 flowing through a wire. The effective value of the resulting current will be:
1. \(15/2~A\)
2.
3.
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 . When only the inductance is removed, the current leads the voltage by . 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 are connected to an AC source as shown. At resonance:
1. | Reading in \(V_3\) = Reading in \(V_1\) |
2. | Reading in \(V_1\) = Reading in \(V_2\) |
3. | Reading in \(V_2\) = Reading in \(V_4\) |
4. | Reading in \(V_2\) = Reading in \(V_3\) |
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/ A
4. 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