Question

A circuit has a section AB shown in the figure. The emf of the source equals $\epsilon =10$V, the capacitance as of the capacitors are equal to $C_1=1.0\mu F$ and $C_2=2.0\mu F$, the potential difference ${\varphi }_A-{\varphi }_B=5.0$V. The voltage across capacitor $C_1$ & $C_2$ is respectively.

• A. $10V,5V$
• B. $10/3V,10/3V$
• C. $0V,0V$
• D. $5/3V,5/3V$

Answer 1

Answer: (A)

Give, e= 10V

$C_1=1\mu F{C}_2=2\mu F$

$V_A-V_B=5V$

$\therefore$ by emf formula,

$V_A-V_B=\frac{q}{C_1}-e+\frac{q}{C_2}$

$V_A-V_B+e=\frac{q(C_1+C_2)}{C_1{C}_2}$

$q=\frac{\left[\right(V_A-V_B)+e]C_1{C}_2}{C_1+C_2}$

Voltage across $C_1$ is $V_1=\frac{q}{C_1}=\frac{\left[\right(V_A-V_B)+e]C_2}{C_1+C_2}$

=$\frac{(5+10)\times 2}{1+2}=10V$

Voltage across $C_2$ is, $V_2=\frac{q}{C_2}=\frac{\left[\right(V_A-V_B)+e]C_1}{C_1+C_2}$

=$\frac{(5+10)\times 1}{1+2}=5V$