In the given circuit, find the potential difference $| V_{A} - V_{B} |$ (in volts) where $V_{A}, V_{B}$ are electric potentials at the points A and B.
$(E_{1} = 1 V, E_{2} = 2 V, E_{3} = 3 V, C_{1} = 1 μF, C_{2} = 2 μF, C_{3} = 3 μF)$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 2.33.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let charges on capacitors be $Q_{1}, Q_{2}, Q_{3}$ respectively
$Q_{1} + Q_{2} + Q_{3} = 0$ (connected plates of $C_{1}, C_{2}, C_{3}$ are isolated)

$V_{A} + E_{1} + \frac{Q_{1}}{C_{1}} = V_{B} \Rightarrow Q_{1} = - C_{1} (V_{A} - V_{B} + E_{1})$

$V_{A} + E_{2} + \frac{Q_{2}}{C_{2}} = V_{B} \Rightarrow Q_{2} = - C_{2} (V_{A} - V_{B} + E_{2})$

$V_{A} + E_{3} + \frac{Q_{3}}{C_{3}} = V_{B} \Rightarrow Q_{3} = - C_{3} (V_{A} - V_{B} + E_{3})$

But $Q_{1} + Q_{2} + Q_{3} = 0$

$- C_{1} (V_{A} - V_{B} + E_{1}) - C_{2} (V_{A} - V_{B} + E_{2}) - C_{3} (V_{A} - V_{B} + E_{3}) = 0$

$1 (V_{A} - V_{B} + 1) + 2 (V_{A} - V_{B} + 2) + 3 (V_{A} - V_{B} + 3) = 0$

$6 (V_{A} - V_{B}) + 14 = 0$

$V_{A} - V_{B} = \frac{- 14}{6} V$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test