Question

For the circuit shown, the value of current at time t=3.2 s will be ________ A. 

 

Hint:

When dealing with time-varying sources from a graph, first determine the analytical expression for the source's value in the relevant time interval. Then apply standard circuit laws like KVL or Ohm's law.

Answer 1

Correct Answer: 1

First, we need to find the value of the voltage source V(t) at the specific time t = 3.2 s from Figure 1.
For the time interval $3 \le t \le 4$, the graph of V(t) is a straight line passing through the points (3, 5) and (4, 10).
The equation of this line is given by $V - V_1 = \frac{V_2 - V_1}{t_2 - t_1}(t - t_1)$.
$V(t) - 5 = \frac{10 - 5}{4 - 3}(t - 3)$.
$V(t) - 5 = 5(t - 3)$.
$V(t) = 5t - 15 + 5 = 5t - 10$.
At $t = 3.2$ s, the voltage is:
$V(3.2) = 5(3.2) - 10 = 16 - 10 = 6$ V.
Now, analyze the circuit in Figure 2 using Kirchhoff's Voltage Law (KVL) for the loop. Let the current be I.
Starting from the V(t) source and moving clockwise:
$V(t) - I \cdot R - 5\text{V} = 0$.
We know $V(t) = 6$ V at $t=3.2$ s and $R = 1 \Omega$.
$6 - I(1) - 5 = 0$.
$1 - I = 0$.
$I = 1$ A.