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(A) From the graph in Figure $1$,for the time interval $t=3 \, s$ to $t=4 \, s$,the voltage $V(t)$ increases linearly from $5 \, V$ to $10 \, V$.The s

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(A) From the graph in Figure $1$,for the time interval $t=3 \, s$ to $t=4 \, s$,the voltage $V(t)$ increases linearly from $5 \, V$ to $10 \, V$.The slope of this line is $m = \frac{10-5}{4-3} = 5 \, V/s$.The equation of the line for $3 \le t \le 4$ is $V(t) - 5 = 5(t - 3)$,which simplifies to $V(t) = 5t - 10$.At $t = 3.2 \, s$,the voltage is $V(3.2) = 5(3.2) - 10 = 16 - 10 = 6 \, V$.From the circuit diagram in Figure $2$,applying Kirchhoff's Voltage Law $(KVL)$ in the loop:$V(t) - iR - 5 = 0$Given $R = 1 \, \Omega$ and $V(t) = 6 \, V$ at $t = 3.2 \, s$:$6 - i(1) - 5 = 0$$i = 1 \, A$.

For the circuit shown,the value of current at time $t=3.2 \, s$ will be ...... $A$. (Voltage distribution $V(t)$ is shown by Figure $1$ and the circuit is shown in Figure $2$)

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