Question

Q.4. Write the result derive by the student in both conditions of parallel and series combination
of capacitor if all the capacitor have different capacitance.​

Answer 1

Answer:

• When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitors' capacitances. If two or more capacitors are connected in parallel, the overall effect is that of a single equivalent capacitor having the sum total of the plate areas of the individual capacitors.

Answer 2

To derive:

Net Capacitance expression in:

• Series combination

• Parallel combination

Derivation:

Series Combination

In series combination of capacitors, the total voltage across the circuit will be equal to the addition of all the voltage across the individual capacitors.

$\therefore \: V = V1 + V2 + .....+Vn$

$= > \dfrac{Q}{C_{net} } = \dfrac{Q}{C1} + \dfrac{Q}{C2} + ....+ \dfrac{Q}{Cn}$

$= > \dfrac{1}{C_{net} } = \dfrac{1}{C1} + \dfrac{1}{C2} + ....+ \dfrac{1}{Cn}$

So, final expression:

$\boxed{ \rm{\dfrac{1}{C_{net} } = \dfrac{1}{C1} + \dfrac{1}{C2} + .... +\dfrac{1}{Cn} }}$

Parallel Combination:

In parallel combination the total charge across the circuit will be equal to the addition of the individual charges across the capacitors.

$\therefore \: Q = Q1 + Q2 + .......+ Qn$

$= > C_{net} V = (C1)V + (C2)V + ....(Cn)V$

$= > C_{net} =C1 + C2 + ..... + Cn$

So, final expression:

$\boxed{ \rm{C_{net} =C1 + C2 + ..... + Cn}}$