Question

Suppose a certain battery has an internal emf of 9.00 V but the potential difference across its terminals is only 80.0 % of that value. Part A If that battery is connected to a 44.0 μF capacitor, how much energy is stored when the capacitor is fully charged? Express your answer with the appropriate units. UEU E = nothing nothing Request Answer

Answer 1

Answer:

$1140.48\times 10^{-6}J$

Explanation:

We have given that the battery has an internal emf of 9 volt

So E = 9 volt

Capacitance $C=44\mu F=44\times 10^{-6}F$

It is given that on the terminal voltage is only 80% of potential difference

So V = 0.8×9 = 7.2 volt

We know that energy stored in the capacitor is given by

$E=\frac{1}{2}CV^2=\frac{1}{2}\times 44\times 10^{-6}\times 7.2^2=1140.48\times 10^{-6}J$