Question

A 5μF capacitor is charged across a 120V battery. It is then disconnected from the

battery and connected to an uncharged capacitor of capacitance 3μF. Find the voltage

across the capacitors​

Answer 1

Answer:

• The voltage across the capacitors is 75 volts

Given:

1. Capacitor (C₁) = 5μF
2. Voltage (V₁) = 120 V
3. Uncharged Capacitor (C₂) = 3μF

Explanation:

$\rule{300}{1.5}$

Here, 5μF capacitor is charged to 120 volt source, and then connected to an uncharged capacitor 3μF. As we know that Charge remains constant, so the charge present in charged 5μF capacitor will be equal to the charge present in both (5μF & 3μF) capacitors when they are connected. So,

$\longrightarrow\sf Q_{1}=Q_{2}\\\\\\\\\longrightarrow\sf C_{1}V_{1}=\bigg(C_{1}+C_{2}\bigg)\times V_{2}$

• Here, V₂ is the common voltage of both capacitors.

$\longrightarrow\sf C_{1}V_{1}=\bigg(C_{1}+C_{2}\bigg)\times V_{2}$

• Substituting the values,

$\longrightarrow\sf 5\mu F\times 120=\bigg(5\mu F+3\mu F\bigg)\times V_{2}\\\\\\\\\longrightarrow\sf 5\mu F\times 120=\bigg(8\mu F\bigg)\times V_{2}\\\\\\\\\longrightarrow\sf \bigg(5\times 10^{-6}\bigg)\times 120=\bigg(8\times 10^{-6}\bigg)\times V_{2}\\\\\\\\\longrightarrow\sf V_{2}=\dfrac{\bigg(5\times 10^{-6}\bigg)\times 120}{8\times 10^{-6}}\\\\\\\\\longrightarrow\sf V_{2}=\dfrac{5\times 120}{8}\\\\\\\\\longrightarrow\sf V_{2}=5\times 15\\\\\\\\\longrightarrow\sf V_{2}=75$

$\longrightarrow\large{\underline{\boxed{\red{\sf V_{2}=75\;volts}}}}$

∴ The voltage across the capacitors is 75 volts.

$\rule{300}{1.5}$