Find the charge and potential difference across C1 of capacitance 1 µF for the given circuit?

Question

Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 2)

Answer 1

Concept:

Calculating the equivalence capacitance, from the left side.

We can see that C₁ and C4 are in parallel

therefore

C_e1 = C₁ + C₄ ....(1)

Now this equivalence will be in series with C2 capacitor

\(\frac{1}{C_{e2}} = \frac{1}{C_{e1}} + \frac{1}{C_1} +\frac{1}{C_3}\) ...(2)

Calculation:

By equation 1,

⇒ C_e1= 1 + 1

\(C_{e1} = 2μ F\)

By equation 2,

⇒ C_e2 = 2/3μF

Thus, the total charge flowing through the circuit

\(Q = C_{eqv}V\)

⇒ Q = 2/3 μF × 4V

⇒ Q = 8/3μC

Charge across will the capacitance will be in the ratio of respective capacitance

C_eqV = C₁V + C₂V + C₃V

Q₁ = Q/2

⇒ Q₁ = 4/3 μC

Now potential difference across C₁ can be calculated as

\(V = \frac{4/3}{C_1}\)

\(⇒ V = \frac{4/3}{1}\)

⇒ V₁ = 4/3V