Question

A 2.37-F and a 5.36-F capacitor are connected in series across a 40.0-V battery. A 8.45-F capacitor is then connected in parallel across the 2.37-F capacitor. Determine the voltage across the 8.45-F capacitor.

Answer 1

Answer:

The voltage across the 8.45F capacitor is 27.73 Volts

Explanation:

Given that,

A  2.37-F and a 5.36-F  capacitor are connected in series across a 40.0 V battery,

$C_{1}=2.37 F\\C_{2}=5.36 F\\V = 40.0 V$

To Find:

V₁  = ? (voltage across the 10.00 µF capacitor)

Solution:

For Capacitor Series Combination we have

$\dfrac{1}{C_{s}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}$

Substituting the values we get

$\dfrac{1}{C_{s}}=\dfrac{1}{2.37}+\dfrac{1}{5.36}\\C_{s}=1.643F$

As Capacitors are connected in series they have the same charge on each plates, and is given by,

$Q= C_{s}\times V$

Substituting the values we get

$Q= 1.643 F\times 40\\=65.72C\\Q=65.72 C$

Also in Series voltage will be different across C₁ and C₂,

Therefore voltage across C₁  will be

$V_{1}= \dfrac{Q}{C_{1}}$

Substituting the values we get

$V_{1}=\dfrac{65.72}{2.37}=27.73\ Volt$

Now according to the given condition,

8.45F capacitor is then connected in parallel across the 2.37 F capacitor,

Therefore we know in Parallel, Voltage remain Same, Hence Voltage across 8.45F capacitor will be same as that of voltage across V₁.

Hence,

The voltage across the 8.45F capacitor is 27.73 Volts