Question

A capacitor of 2uF can withstand a maximum potential difference of 5v. it is connected with a capacitor of 5uF . The series combination can now withstand a potential difference of 7v. So, the maximum voltage 5uF can withstand is.......

Answer 1

Answer:

The two 3μF capacitors in parallel can be combined into a single capacitor (C) with an equivalent capacitance of C=3+3=6μF

Now, the remaining two 3μF capacitors are in series with C. This can be also combined in a single capacitor of capacitance C′.

C′1=31+31+61

C′=56μF

Now, in series charge across all the capacitors is same and voltage add up to give the equivalent voltage.

Let required voltage across the 6μF capacitor be V and the voltage across C′ be V′

Let charge across each of the two capacitors be Q 

Using definition of capacitance and same charge,

Q=Q

6V=56V′

Given V+V′=60

Solving above two equations, V=10 V