Two capacitors of capacitances 4.0 μF and 6.0 μF are connected in series with a battery of 20V. Find the energy supplied by the battery.

Given,

Capacitance are 4.0 μF C₁) and 6.0 μF C₂)

The voltage of battery, V is 20V

Formula used

Let us represent the arrangement as

In a series arrangement the the charge on both the capacitance are same, and can be found out by the equation,

Where Q and V represents the Charge and Potential difference respectively.

in series arrangement with Capacitance C₁ and C₂, C_eff can be found out as,

Or,

The energy stored in the capacitor,E in Jules) can be found out by the relation,

Where C is the capacitance of the capacitor in Farad and V is the potential difference across the capacitor.

We have to find the equivalent capacitance by eqn.2

So,

Or,

Now the energy supplied by the battery is equivalent to the energy stored in the equivalent capacitor with capacitance C_eff. Hence by eqn.3, we get

Or,

The supplied energy will be twice of the stored energy, since half of the supplied energy will be dissipated by the resistance of the circuit.

Hence the supplied energy will be

Hence an amount of 960 μJ will be supplied by the battery.