Alternating Current

When a.c. source is connected across a resistance, the correct phasor relation between the current ($i_R$) and emf ($e_R$) is shown in figure

An alternating voltage given as, $E = 100\sqrt{2}\sin(100t)$ volt is applied to a capacitor of $1\ \mu\text{F}$. The current reading of the ammeter will be equal to

The output of a step-down transformer is measured to be $56$ V when connected to $14$ W light bulb. The value of peak current is (in A)

Consider the statements on capacitive reactance. Statement I: As capacitance increases, the capacitive reactance also increases. Statement II: The capacitive reactance is independent of the source frequency.

An alternating voltage is given as, $E=100\sqrt{2}\sin 100t$ This voltage E is applied to a capacitor of $1\ \mu\text{F}$. The current reading in an a.c. ammeter in mA is

In an AC circuit, E and I are given by $E = 150\sin(150t)$ V and $I = 150\sin\left(150t + \dfrac{\pi}{3}\right)$ A. The power dissipated in the circuit is $\{\cos(60)^\circ = 1/2\}$

If the r.m.s. current in an a.c. circuit is $5$ A, having frequency $50$ Hz. The value of current $\dfrac{1}{300}$ s after its value becomes zero (in A) is $\{\sin(\pi/3) = (3)^{0.5}/2\}$

In an AC circuit, the current is $I = 100 \sin (5t)$ A. The value of $I_{rms}$ is

A step down transformer connected to an a.c. mains of $220$ V is made to operate at $5.5$ V, $44$ W lamp. The current in the primary circuit is (Ignore power losses)

An a.c. source is connected to a capacitor 'C'. Due to increase in its operating frequency

In the LCR series circuit, an alternating e.m.f. $e$ and current $i$ are given by the equation $e = 100\sin(100t)\,V$ and $i = 100\sin\left(100t + \dfrac{\pi}{3}\right)mA$. The average power dissipated in the circuit…

A series LCR circuit is connected to a source of alternating e.m.f. of $50$ V. The potential difference across inductor and capacitor is $80$ V and $50$ V respectively. The potential difference across the resistor is