
The charge $Q$ is given as:
\[ Q = ne = \text{area under } I-t \text{ graph} \]
Substituting the values: \[ n = \frac{5000 \times 10^{-3}}{1.6 \times 10^{-19}} = 3.125 \times 10^{19} \]

The circuit shown in the figure contains two ideal diodes \( D_1 \) and \( D_2 \). If a cell of emf 3V and negligible internal resistance is connected as shown, then the current through \( 70 \, \Omega \) resistance (in amperes) is: 