Question 51 of 180

In the given circuit, the charge on $4 \mu F$ capacitor will be:

A$9.6 \mu C$B$5.4 \mu C$C$24 \mu C$D$13.4 \mu C$

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Answer: (C) $24 \mu C$

The given circuit can be redrawn as:

Charge on $4\mu F$ capacitor:
$Q=\left(\dfrac{4\times 6}{4+6}\right)\times 10$
$Q=24\mu C$
in series combination charge remains same.
so,
${Q}_{4\mu F}={Q}_{6\mu F}=24\mu C$

JEE Main Physics practice question on Capacitors (Grouping of capacitors), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.