Question

A capacitor of capacitance 1μF withstands a maximum voltage of 6kV, while another capacitor of capacitance 2μF, the maximum voltage 4kV. If they are connected in series, the combination can withstand a maximum of

• A. (a) 6kV
• A. (b) 4kV
• A. (c) 10kV
• A. (d) 9kV

Answer 1

(d)

When the two condensers are connected in series.

$C = \frac{2\text{ x 1}}{\text{2 } + 1} = \frac{2}{3}$μF and $Q = \frac{2}{3}E$

The potential of condenser $C_{1}$is given by

$V_{1} = \frac{Q}{C_{1}} = \frac{2}{3}E < 6kV$

$\Rightarrow \mathrm { E } < 9 \mathrm { kv }$ $v_{2} = \frac{Q}{c_{2}} = \frac{E}{3} < E < 12KV$

To avoid break down E$\leq$9KV