JEE MAINS PRACTICE · Physics
Alternating Current
638 practice questions for JEE Mains Practice — 270 easy · 238 medium · 130 hard. Every question is graded instantly with a step-by-step solution when you miss it.
Start free practice →Sample questions
MCQ
Which phasor diagram represents LCR circuit at resonance?
MCQ
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
MCQ
At resonance, the LC parallel resonant circuit
MCQ
When a $60$ mH inductor and resistor are connected in series with an AC source, the voltage leads the current by $60^\circ$. If the inductor is replaced by a $0.5\ \mu$F capacitor, the voltage lags behind the current by…
MCQ
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
MCQ
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
MCQ
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.
MCQ
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 $\{\sin(\pi/3) = (3)^{0.5}/2\}$
MCQ
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
MCQ
An a.c. source is connected to a capacitor 'C'. Due to increase in its operating frequency
MCQ
The r.m.s value of current in a 50 Hz A.C. circuit is 10 A. The average value of A.C. current over a cycle is
MCQ
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 $\{\cos(60)^\circ = 1/2\}$