A LED has a voltage drop of 2V across it and passes a current of 10 mA. When it operates with a 6V battery through a limiting resistor R. The value of R is
40 kΩ
200 Ω
4 kΩ
400 Ω
A series R-L-C circuit driven with Erms = 100V and frequency fd = 50 Hz is shown below.
What new capacitance should be connected to maximize the average power, if other parameters of the circuit are not changed ?
41.6 µF
33.5 µF
39.8 µF
31.3 µF
An L-C-R series circuit with 100 Ω resistance is connected to an AC source of 200 V and angular frequency 300 rad s-1 . 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°. The average power dissipated in this L-C-R circuit is
0
400 W
200 W
800 W
Given,
R1 = 10 Ω, C1 = 2µF, R2 = 20 and C2 = 4F
The time constants (in µs) for the circuits I, II, III are respectively.
18, 18/9, 4
4, 8/9, 18
18, 4, 8/9
8/9, 18, 4
In the given circuit, the AC source has ω = 100 rad/s, considering the inductor and capacitor to be ideal,
the current through the circuit is 0.4 A
the current through the circuit is 0.3 A
the voltage across 100 Ω resistor is 10 V
the voltage across 50 Ω resistor is 10 V
The impedence of a circuit, when a resistance R and an inductor of inductance L are connected in series in an AC circuit of frequency f, is
In a series LCR circuit, resistance R = 10 Ω and the impedance Z = 10 Ω. The phase difference between the current and the voltage is
0°
30°
45°
60°
A circuit consists of an inductance of 0.5 mH and a capacitor of 20 µF. The frequency of the LC oscillations is approximately
400 Hz
88 Hz
1600 Hz
2400 Hz
The maximum value of AC voltage in a circuit is 707 V. Its rms value is
70.7 V
100 V
500 V
707 V