Search any question & find its solution
Question:
Answered & Verified by Expert
In the given circuit, a charge of $+80 \mu C$ is given to the upper plate of the $4 \mu F$ capacitor. Then in the steady state, the charge on the upper plate of the $3 \mu F$ capacitor is
Options:
Solution:
2806 Upvotes
Verified Answer
The correct answer is:
$+48 \mu \mathrm{C}$
The total charge on plate $A$ will be $80 \mu \mathrm{C}$.
$2 \mu F$ and $3 \mu F$ capacitors are in parallel. Therefore, $C_{e q}=2+3=5 \mathrm{HF}$
Charge on capacitor of $3 \mu \mathrm{F}$ capacitance $q=\frac{3}{5} \times 80=48 \mu C$
$2 \mu F$ and $3 \mu F$ capacitors are in parallel. Therefore, $C_{e q}=2+3=5 \mathrm{HF}$
Charge on capacitor of $3 \mu \mathrm{F}$ capacitance $q=\frac{3}{5} \times 80=48 \mu C$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.