In an a.c. circuit with pure capacitance $C$ and a.c. source $E = E_0 \sin \omega t$, the equation of instantaneous current is given by
- A$I = E_0 \omega C \sin(\omega t)$
- B$I = E_0 \omega C \sin(\omega t + \pi/2)$
- C$I = \frac{E_0}{\omega C} \sin(\omega t)$
- D$I = \frac{E_0}{\omega C} \sin(\omega t + \pi/2)$
Explore More
Similar Questions
The same current is flowing in two alternating circuits. The first circuit contains only an inductor and the other contains only a capacitor. If the frequency of the alternating e.m.f. is increased,the values of current will
$A$ parallel plate capacitor having plates of radius $6 \text{ cm}$ has capacitance $100 \text{ pF}$. It is connected to a $230 \text{ V}$ a.c. supply with angular frequency $300 \text{ rad/s}$. The r.m.s. value of current is:
$A$ capacitor of capacitance $10 \mu F$ is connected to an $AC$ source and an $AC$ Ammeter. If the source voltage varies as $V = 50 \sqrt{2} \sin 100 t$, the reading of the ammeter is (in $\text{ mA}$)
MediumKCET 2016
View SolutionAn alternating $emf$ $E = 440 \sin(100 \pi t)$ is applied to a circuit containing an inductance of $\frac{\sqrt{2}}{\pi} \text{ H}$. If an $AC$ ammeter is connected in the circuit,its reading will be $....... \text{ A}$.
DifficultJEE MAIN 2022
View SolutionWhen an $A$.$C$. source is connected across a pure capacitor,the correct phase relation between current $(i_c)$ and voltage $(e_c)$ is shown in the figure.
DifficultMHT CET 2022
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo