A 4μF capacitor is charged to 400 volts and then its plates are joined through a resistance of 1 kΩ. The heat produced in the resistance is:

Concept:

When the capacitor discharges through the resistor, all the stored energy will be converted into heat. Thus, the heat produced in the resistor will be equal to the energy initially stored in the capacitor.

• The capacitance of a capacitor (C): The capacity of a capacitor to store the electric charge is called capacitance.
  • The capacitance of a conductor is the ratio of charge (Q) to it by a rise in its potential (V).
  • C = Q/V
• The unit of capacitance is the farad, (symbol F).
• Energy stored (U) in the capacitor is given by:

\(U = \frac{1}{2}C{V^2} = \frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2}\;QV\)

Where Q = charge stored on the capacitor, U = energy stored in the capacitor, C = capacitance of the capacitor and V = Electric potential difference

Calculation:

Given,

C = 4 μF = 4 × 10-6 F

Voltage applied = 400 V

Energy Stored  = Heat energy produced through resistance.

\(U = \frac{1}{2}C{V^2} = \frac{1}{2} × {(4 × 10^{-6})} × {(400^2)}\)

⇒ Heat Produced = 32 × 10^-2 Joule

⇒ Heat Produced = 0.32 Joule.