Question
A parallel plate capacitor is charged with a potential difference 50 V . Thereafter it is discharged by a resistor for 2 sec , due to which its potential decreases by 10 V . Calculate the portion of energy stored in the capacitor.
✓
Answer
Initial energy stored in the capacitor
$ \begin{aligned} U_i & =\frac{1}{2} CV^2=\frac{1}{2} \times C(50)^2 \\ & =\frac{1}{2} C(50)^2 \end{aligned} $
After 2 s , when the potential decreases by 10 V , then the final potential will be 40 V .
Final energy stored in the capacitor $ U_f=\frac{1}{2} \times C(50)^2 $
Part of stored energy $=\frac{ U _f}{ U _i}$
$=\frac{\frac{1}{2} \times C (40)^2}{\frac{1}{2} \times C (50)^2}=\frac{40^2}{50^2}$
$=\frac{1600}{2500}=\frac{16}{25}$
$=0.64$ Ans.
$ \begin{aligned} U_i & =\frac{1}{2} CV^2=\frac{1}{2} \times C(50)^2 \\ & =\frac{1}{2} C(50)^2 \end{aligned} $
After 2 s , when the potential decreases by 10 V , then the final potential will be 40 V .
Final energy stored in the capacitor $ U_f=\frac{1}{2} \times C(50)^2 $
Part of stored energy $=\frac{ U _f}{ U _i}$
$=\frac{\frac{1}{2} \times C (40)^2}{\frac{1}{2} \times C (50)^2}=\frac{40^2}{50^2}$
$=\frac{1600}{2500}=\frac{16}{25}$
$=0.64$ Ans.
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