CUET Preparation Today
The charge on capacitor of capacitance 20 μF as shown in figure given below is:
|
66.7 μC 100 μC 300 μC 450 μC |
100 μC |
The correct answer is Option (2) → 100 μC As capacitor are connected in series - $\frac{1}{C_{net}}=\frac{1}{20}+\frac{1}{10}=\frac{1+2}{20}$ $C_{net}=\frac{20}{3}μF$ Now, $C=\frac{Q}{V}$ $⇒Q=CV$ $=\frac{20}{3}×15=100 μC$ |
