Related InterviewSolutions

• Figure shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance `1Omega//m`. Position of the conducting rod at `t=0` is shown. A time dependent magnetic field `B=2t` tesla is switched on at `t=0` At `t=0`, when the magnetic field is switched on, the conducting rod is moved to the left at constant speed `5cm//s` by some external means. At `t=2s`, net induced emf has magnitudeA. (a) `0.12 V`B. (b) `0.08 V`C. ( c) `0.04 V`D. (d) `0.02 V`
• The unit of `L//R` is (where `L=` inductance and `R=` resistance)A. AmpereB. VoltC. per secD. sec
• A series LCR circuit having `L = 10 mH`, `C = (400//pi^(2)) mu F` and R = 55 ohm is connected to 220 v variable frequency a.c. supply. (i) Find frequency of source, for which average power absorbed by the circuit is maximum (ii) Calculate the amplitude of current.
• The household supply voltage as measured by an a.c. voltmeter is 200 meter is 220 volts. If the frequency of a.c. Supply is 50 Hz, then the equation of the line voltage, will beA. 1. `V=220 sin (100 pi t)`B. 2.`V=110 sin (50 pi t)`C. 3.`V=440 sin (100 pi t)`D. `4.V=311 sin (100 pi t)`
• The domestic power supply of 220 V, 50 Hz is connected to a resistor. What is the time taken by the alternating current flowing in the resistor, to change from its maximum value to turns value?A. `5 xx 10^(-3) s`B. `2.5 xx 10^(-3) S`C. `10 xx 10^(-3) S`D. `2.5 xx 10^(3)S`
• A 200 ohm electric iron is connected to 220 V, 50 Hz a.c. supply. Calculate the average power delivered to iron.
• The primary winding of a transfomer has 50 turns while its secondary has 500 turns. If the primary is connected to an a.c. supply of 220 V, 50 Hz, then the output at the secondaray will beA. 220V, 50HzB. 2200V, 50HzC. 2200V, 500HzD. 22V, 5Hz
• In India, domestic power suppy is at 220 V, 50 hz, while in U.S.A, its 110 V, 60 hz. Give one advantage and one disadvantage of 220 V supply over 110 V supply.
• An e.m.f. `E =E_(0) cos omega t` is applied to a circuit containing L and R in series. If `X_(L)=R`, then the power dissipated in the circuit is given byA. `(E_(0)^(2))/(8R)`B. `(E_(0)^(2))/(4R)`C. `(E_(0)^(2))/(2R)`D. `(E_(0)^(2))/(R)`
• Alternating current through an inductor lags behind the alternating voltage by `90^(@)`. What does it imply ?