For the `AC` circuit shown, the reading of ammeter and voltmeter are `5A` and `50 sqrt(5)` volts respectively, then A. average power deliverd by the source is `250 W`B. rms value of `AC` source is 50 voltsC. voltage gain is 2D. frequency of ac source is 1000rad/s

For the `AC` circuit shown, the reading of ammeter and voltmeter are `

1.	For the `AC` circuit shown, the reading of ammeter and voltmeter are `5A` and `50 sqrt(5)` volts respectively, then A. average power deliverd by the source is `250 W`B. rms value of `AC` source is 50 voltsC. voltage gain is 2D. frequency of ac source is 1000rad/s
Answer» Correct Answer - A,B,C,D `i_("rms") = 5A` Reading of voltmeter `= i_("rms") sqrt(R^(2) + X_(C )^(2))` From this `X_(C ) = 20 Omega implies omega = 10^(3) "rad"//s` `:. X_(L) = omega L = 20 Omega` Therefore the circuit is in resonance `E_("rms") - i_("rms") R = 50 V` also power factor = 1 Av, power `= E_("rms") i_("rms") cos phi = 250 W` voltage gain ` = (X_(L))/(R ) = 2`

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

For the `AC` circuit shown, the reading of ammeter and voltmeter are `5A` and `50 sqrt(5)` volts respectively, then A. average power deliverd by the source is `250 W`B. rms value of `AC` source is 50 voltsC. voltage gain is 2D. frequency of ac source is 1000rad/s

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment