Question

For the circuit shown, the value of current at time $t=3.2\text{s}$ will be
$\text{(in A)}$.



[Voltage distribution $V\left(t\right)$ is shown by Figure $1$ and the circuit is shown in Figure $2$]

Answer 1

From graph we can say that voltage is increasing linearly from $t=3\text{s}\text{to}4\text{s}$. So, slope is given as $\frac{\Delta V}{\Delta t}=5$.
So, voltage at $t=3.2\text{s}$ is equal to $5+0.2\times 5=6\text{V}$


Therefore, current at the given time $t=3.2\text{s}$, will be
$i=\frac{6-5}{1}=1\text{A}$