Alternating Current
Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.
A steady potential difference of 100 V produces heat at a constant rate in a resistor. The alternating voltage which will produce half the heating effect in the same resister will be
The power supplied to the resistor by DC source $$=\dfrac{V_{dc}^2}{R}$$
Energy given by AC source $$=\int_0^T \dfrac{V_0^2}{R}dt$$
Hence, $$\int_0^T \dfrac{V_0^2sin^2\omega t}{R}dt=\dfrac{1}{2}\times \dfrac{V_{dc}^2T}{R}$$
$$\implies V_0=V_{dc}=100V$$
Find the time required for a 50 Hz alternating current to change its value from zero to the rms value.
The household supply of electricity is at 220 V (rms value) and 50 Hz. Find the peak voltage and the least possible time in which can change from the rms value to zero.
The maximum value of a.c. voltage in a circuit is 707V. Its r.m.s. value is
What is the $$r.m.s.$$ value of an alternating current which when passed through a resistor produces head which is thrice of that produced by a direct current of $$2$$ amperes in the same resistor:-
If $$ E_0 $$ represents the peak value of the voltage in an ac circuit, the r.m.s. value of the voltage will be[
The ratio of peak value and r.m.s value of an alternating current is
A mixer of 100$$\Omega $$ resistance is connected to an A.C. source of 200V and 50 cycles/sec. The value of average potential difference across the mixer will be:
$$\displaystyle I = 6\cos wt + 8\sin wt$$ is applied across a resistor of 40 $$\Omega$$. Find the potential difference across the resistor.
The instantaneous voltage from an ac source is given by $$E=300 sin 314t$$; what is the rms voltage of the source?
A light bulb is rated at $$100W$$ for a $$220V$$ ac supply. Calculate the resistance of the bulb.