Question

Two capacitors of 3 µF and 6 µF are connected in series with a battery of P. d. 12 V. The P. d. across 3µF and 6µF capacitors respectively will be​

Answer 1

Question:-

Two capacitors of 3 µF and 6 µF are connected in series with a battery of P. d. 12 V. The P. d. across 3µF and 6µF capacitors respectively will be.

Given that;

• ${\sf{{C_{1}}}}{\sf{{ = 3µF, _{}}}}$
• ${\sf{{C_{2}}}}{\sf{{ = 6µF, _{}}}}$
• ${\sf{{V=12Volt_{}}}}$

To Find:-

The potential difference across 3µF and 6µF capacitors.

Solution:-

Since both the capacitors are 6µF capacitors.

${\sf{{So;C_{net}}}} = \frac{{\sf{{{{\sf{{C_{} }}}} _{1}}}} \times {\sf{{C_{2}}}}}{{\sf{{C_{1} + }{{\sf{{C_{2} }}}} }}} = \frac{3 \times 6}{3 + 9} = 2µF$

So; the capacitors are connected in series,hence the charge in both capacitors are same.

${\sf{{Q_{net}}}}{\sf{{(total \: charge)_{}}} }{\sf{{ \: = \: C_{net}}} } \times{\sf{{(P.d)_{}}}}$

${\sf{{So; Q_{ net}}}}{\sf{{=2µF \times 12 = 24×10 {}^{ - 6} coloum_{}}}}$

So; the potential difference across 3µF are;

${\sf{{Q_{net}}}}{\sf{{ \: = \: C_{1}}}}{\sf{{ V_{1}}}}$

${\sf{{So;V_{1}}}} = \frac{{\sf{{Q_{net}}}}}{{\sf{{G_{}}}}} = \frac{24 \times 10 {}^{ - 6} }{3 \times 10 {}^{ - 6} } = {\sf{{ 8Volts._{}}}}$

Since; total potential difference = P.d across 3µF + Potential drop across 6µF .

${\sf{{So; potential \: drop \: across \: 6µF \: _{}}}} \\ {\sf{{ = V \: – \: V_{1}}}}{\sf{{ \: = \: 12 - 8 = 4Volts. _{}}}}$

Hence,8V,4V is the are the potential difference across 3µF and 6µF capacitors.

Answer 2

Given:

Two capacitors of 3 µF and 6 µF are connected in series with a battery of P. d. 12 V.

To find:

Potential difference across the capacitors ?

Calculation:

First we find the net capacitance:

• Net capacitance = (3×6)/(3+6) = 18/9 = 2µF.

Now net charge across the capacitors will be :

$= C_{net} \times V$

$= 2 \times 12 = 24 \: \mu C$

So, potential difference across :

• 3µF will be = 24/3 = 8 Volts.

• 6µF will be = 24/6 = 4 Volts.

Hope It Helps.