A combination of elements is enclosed in a black box an the voltage and currents are measured across this black box. The expression for applied voltage, and the current flowing in it is `V = V_(0) sin omega t`, `i = 2 sqrt(2) sin (omega + pi//4)` where `omega = 100 pi "rad"//"sec"` Then the wrong statement is A. The must be a capacitor is the black boxB. Power factor of circuit = 0.707C. There must be a resistor in the boxD. There must be an inductor in the box

A combination of elements is enclosed in a black box an the voltage an

1.	A combination of elements is enclosed in a black box an the voltage and currents are measured across this black box. The expression for applied voltage, and the current flowing in it is `V = V_(0) sin omega t`, `i = 2 sqrt(2) sin (omega + pi//4)` where `omega = 100 pi "rad"//"sec"` Then the wrong statement is A. The must be a capacitor is the black boxB. Power factor of circuit = 0.707C. There must be a resistor in the boxD. There must be an inductor in the box
Answer» Correct Answer - D Since current leads voltage by `45^(@)`, there mustbe a resistor and a capacitor. We can say nothing about inductor surely. It may or may not be present.

Discussion

No Comment Found

Related InterviewSolutions

Can the phenomenon of resonance be exhibited in RL or RC circuit.
The impedance of a sereis `RL` circuit is same as the series `RC` circuit when connected to the same `AC` source separately keeping the same resistance. The frequency of the source isA. `(1)/(sqrt(LC))`B. `(1)/(2pi sqrt(LC))`C. `(R )/(L)`D. `(1)/(RC)`
The equation of an alternating voltage is `E = 220 sin (omega t + pi // 6)` and the equation of the current in the circuit is `I = 10 sin (omega t - pi // 6)`. Then the impedance of the circuit isA. 10 ohmB. 22 ohmC. 11 ohmD. 17 ohm
In an ac circuit, the current lags behind the voltage by `pi//3`. The components in the circuit areA. R and LB. R and CC. L and CD. Only R
In an ac circuit, the current lags behind the voltage by `pi//3`. The components in the circuit areA. R and LB. L and CC. R and CD. only R
A 200 km telephone wire has capacity of `0.014 muF km^(-1)`. If it carries an alternating current oif frequency 50kHz, what should be the value of an inductance required to be connected in series so that impedance is minimumn ?
In the circuit shown in figureure the `AC` source gives a voltage `V=20cos(2000t)`. Neglecting source resistance, the voltmeter and and ammeter readings will be .A. `0V,2.0A`B. `0V,1.4A`C. `5.6V,1.4A`D. `8V,2.0A`
A signal generator supplies a sine wave of `200V, 5kHz` to the circuit shown in the figureure. Then, choose the wrong statement. .A. The current in te resistive brance is `0.2 A`B. the current in the capacitive branch is `0.126 A`C. Total line current is `~~0.283A`D. Current in both the branches is same
In an `AC` circuit, the applied potential difference and the current flowing are given by `V=20sin100tvolt,I=5sin(100t-pi/2)` amp The power consumption is equal toA. `1000W`B. `40W`C. `20W`D. zero
In an `AC` circuit, the current is given by `i=5sin(100t-(pi)/2)` and the `AC` potential is `V=200sin(100t)`volt. Then the power consumption isA. `20` wattsB. `40` wattC. `1000` wattD. `0` watt

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment