In the given circuit the potential difference across the $7 μ F$ capacitor is $6 V$ . Then

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

the charge on the $3 μ F$ capacitor is $42 μ C$

the emf of the battery is $30 V$

the potential difference across the $12 μ F$ capacitor is $10 V$

the potential difference across the $3.9 μ F$ capacitor is $10 V$

answer is A, C, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

Since same charge flows through $7 μ F$ and $3 μ F$ (as both are in series), so

$Q_{7} = Q_{3} = 42 μ F \Rightarrow 7 \times 6 = 3 V_{3} V_{3} = 14 V$

So $V_{3.9} = 20 V$

Charge on $12 μ F$ capacitor is

$Q_{12} = Q_{3.9} + Q_{7} Q_{12} = 3.9 \times 20 + \frac{7 \times 3}{7 + 30} \times 20$

$Q_{12} = 120 μ C$ $=$ Total charge in the circuit

$V_{12} = \frac{120}{12} = 10 V$

Voltage across battery is $14 + 6 + 10 = 30 V$

JEE

NEET

Foundation JEE

Foundation NEET

CBSE