Question

58. Two condenser of 8uF and 4uF
are connected in series. The p.d.
across them is 1200 volt. The p.d.
across 8uF is-​

Answer 1

Answer:

400 volt

Explanation:

two capacitors C1 = 8uF and C2 = 4uF are connected in series so, equivalent capacitance is Ceq = C1.C2/(C1 + C2) = 8 × 4/(8 + 4) = 32/12 uF

we know,

Q = CV

where Q is charge , C is the capacitance of capacitor and V is the potential difference.

now, Qnet in circuit = equivalent capacitance of capacitors attached in circuit × potential difference

Qnet = 1200 × 32/12 = 3200uC

both the capacitors are in series so, charge passing through both are same. e.g., Q = 3200uC

hence, Q = CV

here, C = 8uF , Q = 3200 uC

now, V = 3200uC/8uF = 400v

hence, potential difference across 8uF is 400v