Using an ac voltmeter, the potential difference in the electrical line in a house is read to be 234 V. If the line freqency is known to be 50 cycles per second, the equation for the line voltage isA. `V = 165 sin (100 pi t)`B. `V = 331 sin (100 pi t)`C. `V = 220 sin (100 pi t)`D. `V = 440 sin (100 pi t)`

Using an ac voltmeter, the potential difference in the electrical line

1.	Using an ac voltmeter, the potential difference in the electrical line in a house is read to be 234 V. If the line freqency is known to be 50 cycles per second, the equation for the line voltage isA. `V = 165 sin (100 pi t)`B. `V = 331 sin (100 pi t)`C. `V = 220 sin (100 pi t)`D. `V = 440 sin (100 pi t)`
Answer» Correct Answer - 2 `E = E_(0) sin omega t`, voltage read is r.m.s value `E_(0) = sqrt(2) xx 234 V = 331` volt and `omega t = 2 pi nt = 2 pi xx 50 xx t = 100 pi t` Thus, the eqn of line voltage is given by `V = 331 sin (100 pi t)`

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