
For this interval, the current is constant, so the area under the curve is simply a rectangle:
\[ Q_1 = I \times \Delta t = 2 \times (3 - 1) = 4 \, \text{C} \]
For this interval, the current-time graph forms a triangle. The area of a triangle is given by:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Substituting the values (base = 2 s, height = 2 A):
\[ Q_2 = \frac{1}{2} \times (6 - 4) \times 2 = 2 \, \text{C} \]
From the calculations above:
Thus, we have: \[ Q_1 > Q_2 \]
Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.


For the circuit shown above, the equivalent gate is:
