In the circuit below, the capacitors are fully charged. What is the voltage across C_{1}, C_{2}, and C_{3} respectively?

This question was previously asked in

VSSC (ISRO) Technician B (Electronic Mechanic) Previous Year Paper (Held on 25 Sept 2016)

Option 4 : 2.5 V, 2.5 V, 5 V

Subject Test 1: Electrical Basics

3012

20 Questions
80 Marks
30 Mins

__Concept__:

The capacitor acts as an open circuit to DC under steady-state.

The voltage stored by a capacitor C1 connected in series with capacitor C2 is given by:

\(V_{C1}=V× \frac{C_2}{C_1+C_2}\)

__Note__: The above formula is similar to the current division rule used in finding the current through a branch when 2 branches are in parallel.

__Calculation__:

For the given data, the voltage across the first 4 F capacitor will be:

\(V_{C1}=10× \frac{4/3}{4/3+4}\)

(4/3 is the parallel equivalent of opposite resistances 4 and 2)

**V _{C1} = 2.5 V**

Similarly,

\(V_{C2}=10× \frac{4/3}{4/3+4}\)

**VC2 = 2.5 V**

Since the input voltage is 10 V, the voltage across the 2F capacitor will be:

V_{C3} = 10 - (2.5 + 2.5) = 5V

We can also cross-check this as:

\(V_{2 F}=10× \frac{2F}{2F+2 F}\)

V4μF = **5V**