A circuit has a section AB as shown in the fi | vedclass

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

Question

(D) Let $V_A$ and $V_B$ be the potentials at points $A$ and $B$ respectively. Given $V_A - V_B = V$.Let $q$ be the charge on the capacitors. Since the

Vedclass · Accepted Answer

$(V-E)C_2/(C_1+C_2)$

Vedclass · Answer

(D) Let $V_A$ and $V_B$ be the potentials at points $A$ and $B$ respectively. Given $V_A - V_B = V$.Let $q$ be the charge on the capacitors. Since they are in series,the charge $q$ is the same for both.The potential difference across $C_1$ is $V_1 = q/C_1$ and across $C_2$ is $V_2 = q/C_2$.Applying Kirchhoff's Voltage Law from $A$ to $B$:$V_A - V_1 - E - V_2 = V_B$$V_A - V_B = V_1 + V_2 + E$$V = q/C_1 + q/C_2 + E$$V - E = q(1/C_1 + 1/C_2) = q(C_1 + C_2)/(C_1 C_2)$$q = (V - E) (C_1 C_2) / (C_1 + C_2)$The potential difference across $C_1$ is $V_1 = q/C_1 = [(V - E) (C_1 C_2) / (C_1 + C_2)] / C_1 = (V - E) C_2 / (C_1 + C_2)$.

$A$ circuit has a section $AB$ as shown in the figure. If the potential difference between points $A$ and $B$ is $V$ volts,then the potential difference across $C_1$ is

Similar Questions

Find the equivalent capacitance across points $A$ and $B$.

To obtain $3\,\mu F$ capacitance from three capacitors of $2\,\mu F$ each,they will be arranged as:

Three capacitors of $1\,\mu F$,$2\,\mu F$,and $4\,\mu F$ are connected in series to a $10\,V$ source. The charge on the plates of the middle capacitor is:

$A$ network of four capacitors with capacitances $C_1 = C$,$C_2 = 2C$,$C_3 = 3C$,and $C_4 = 4C$ is connected to a battery as shown in the figure. The ratio of the charges on $C_2$ and $C_4$ is

Two capacitors of $10\,\mu F$ and $20\,\mu F$ are connected in series with a $30\,V$ battery. The charge on the capacitors will be,respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module