Question

(i) Calculate the value of $C$.

(ii) Calculate the charge on each capacitor.

(iii) What will be the potential drop across each capacitor?

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or $4=20+C20C $

or $80+4C=20C$ or $C=80/16=5μF$

ii) The equivalent charge $Q_{eq}=C_{eq}V=4(12)=48μC$

As the capacitors are in series so they have same charge is equal to $Q_{eq}$

Thus, $Q_{1}=Q_{2}=48μC$

iii) Potential across $20μF$ is $V_{1}=Q1/C_{1}=48/20=2.4V$

and potential across $C=5μF$ is $V_{2}=Q_{2}/C_{2}=48/5=9.6V$

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