The output E of an alternating voltage s...

The output E of an alternating voltage supply of frequency 50 Hz is shown in Fig. From this Fig, state
(i) value of time `t_(1)`
(ii) peak voltage `E_(0)` (iii) root mean square voltage `E_(v)`
(iv) mean supply is connected in series with a resistance of `2.4 Omega`, calculate the mean power dissipated in the resistor.

Similar Questions

Explore conceptually related problems

The output E of an alternating voltage supply of frequency 50 Hz is shown in Fig. From this Fig, state root mean square voltage E_(v)

The variation with time t of the output E of an alternating voltage supply of frequency 50 Hz is shown in the figure. State the time t_1 , the peak value E_0 of the voltage, the root mean square voltage E_(rms) , the mean (Average) voltage E_(av)

The variation with time t of the output E of an alternating voltage supply of frequency 50 Hz is shown in the figure. The alternating supply is connected in series with a resistor of resistance 2.4 ohm. Calculate the mean power dissipated in the resistor.

A 31.4Omega resistor and 0.1H inductor are connected in series to a 200V, 50Hz ac source. Calculate (ii) the voltage (rms) across the inductor and the resistor.

Expression for instantaneous value of an alternating voltage is given below: epsilon= 150 sin 200 pit where t is the time in seconds . Calculate (i) the peak value of voltage (ii) rms value of voltage and (iii) frequency of supply.

An alternating voltage given by e=140 sin 3142t is connected across a pure resistor of 50Omega .Find:- the frequency of the source

A Sinusoidal voltage V = 200 sin 314 t r is applied to a resistor of 10 ohm. Calculate (i) rms value of voltage (ii) rms current (iii) power dissipated as heat.

An alternating voltage is given by: e = e_(1) sin omega t + e_(2) cos omega t . Then the root mean square value of voltage is given by:

Similar Questions

Recommended Questions