Question

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

• A. $2.5kV$
• B. $10/3kV$
• C. $3kV$
• D. $1kV$

Answer 1

Answer: (A)

$2.5kV$

At above 3 and 2 capacitors are in series so they have same charge. Minimum charge for $3C$ is $Q=3\left(1\right)=3kC$. Thus we can not apply charge greater than $3kC$. For this charge potential across $2C$ is $V=3/2=1.5kV$. Total potential difference between 3C and 2C is $V_t=1+1.5=2.5kV$
Similarly for below branch 7C and 3C , $Q_{min}=6kC$ and potential across 7C is $V=6/7kV$ .Total potential difference between 7C and 3C is $V_t=6/7+2=2.85kV$
As above and below branch are in parallel so its has common potential . Thus the maximum emf should be $2.5kV$