Question

The potential difference between points B and E of the circuits is ______.

 

Options

• ${\displaystyle \frac{(C_2-C_1)}{V}}$

• ${\displaystyle \frac{(C_4-C_3)}{V}}$

• ${\displaystyle \left\{\frac{C_2{C}_3-C_1{C}_4}{C_1+C_2+C_3+C_4}\right\}V}$

• ${\displaystyle \left\{\frac{C_1{C}_4-C_3{C}_2}{(C_1+C_2)(C_3+C_4)}\right\}V}$

Answer 1

The potential difference between points B and E of the circuits is ${\displaystyle \underline{\left\{\frac{C_1{C}_4-C_3{C}_2}{(C_1+C_2)(C_3+C_4)}\right\}V}}$.

Explanation:

Charge on C₁ and C₂

${\displaystyle Q_1=Q_2=\left(\frac{C_1{C}_2}{C_1+C_2}\right)V}$

and charge on C₃ and C₄

${\displaystyle Q_3=Q_4=\left(\frac{C_3{C}_4}{C_3+C_4}\right)V}$

So, ${\displaystyle V_B-V_E=\frac{Q_2}{C_2}-\frac{Q_4}{C_4}}$

= ${\displaystyle \frac{C_{1}V}{C_1+C_2}-\frac{C_{3}V}{C_3+C_4}}$

= ${\displaystyle \left\{\frac{C_1{C}_4-C_2{C}_3}{(C_1+C_2)(C_3+C_4)}\right\}V}$