Question

The diagram shows four capacitors with capacitance and break down voltages as mentioned. What should be the maximum value of the external emf source (in $kV$) such that no capacitor breaks down?

Answer 1

By voltage division formula in upper branch ,
$V_1=\frac{2C}{3C+2C}ϵ=2ϵ/5$
$V_2=\frac{3C}{3C+2C}ϵ=3ϵ/5$
By voltage division formula in lower branch
$V_3=\frac{3C}{3C+7C}ϵ=3ϵ/10$
$V_4=\frac{7C}{3C+7C}ϵ=7ϵ/10$
Equating voltage across each capacitor to its respective breakdown voltage

$V_1=2ϵ/5=1kV$ or $ϵ=2.5kV$
$V_2=3ϵ/5=1kV$ or $ϵ=3.3kV$
$V_3=3ϵ/10=1kV$ or $ϵ=3.3kV$
$V_4=7ϵ/10=1kV$ or $ϵ=2.8kV$
Maximum safe voltage for supply is the minimum ε amongst above values
i.e.
$ϵ=2.5kV$
Thus, safer voltage of the circuit is $2.5kV$