When an A.C. source is connected across a pur | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For a pure capacitor connected to an $A$.$C$. source,the voltage is given by $e_c = V_0 \sin(\omega t)$.The charge on the capacitor is $q = C e_c

Vedclass · Accepted Answer

$(B)$

Vedclass · Answer

(B) For a pure capacitor connected to an $A$.$C$. source,the voltage is given by $e_c = V_0 \sin(\omega t)$.The charge on the capacitor is $q = C e_c = C V_0 \sin(\omega t)$.The current in the circuit is $i_c = \frac{dq}{dt} = \frac{d}{dt} [C V_0 \sin(\omega t)] = \omega C V_0 \cos(\omega t) = \omega C V_0 \sin(\omega t + \frac{\pi}{2})$.This shows that the current $i_c$ leads the voltage $e_c$ by a phase angle of $\frac{\pi}{2}$ radians.In the given phasor diagrams,the vector representing $i_c$ should be $90^{\circ}$ ahead of the vector representing $e_c$ in the counter-clockwise direction.Looking at the options,in diagram $(B)$,the current vector $i_c$ is $90^{\circ}$ ahead of the voltage vector $e_c$.Therefore,the correct option is $(B)$.

When 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.

Similar Questions

The average power dissipated in a pure inductor is

In a purely inductive circuit,the current is

$A$ lamp is connected in series with a capacitor. Predict your observations for $dc$ and $ac$ connections. What happens in each case if the capacitance of the capacitor is reduced?

An $A.C.$ source is connected across a pure inductor. Which one of the following figures shows the correct phase relation between the current and $e.m.f.$?

An alternating voltage $E = 200 \sqrt{2} \sin(100 t) \text{ V}$ is connected to a $1 \mu\text{F}$ capacitor through an $A.C.$ ammeter. The reading of the ammeter shall be: (in $\text{ mA}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module