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

[Voltage distribution V(t) is shown by Fig. (1) and the circuit is shown in Fig. (2)]

We need to determine the value of the electric current flowing through the circuit at the specific time $$t = 3.2\text{ s}$$.

1. Analyze the Voltage Distribution Curve

From the question details , the variable voltage function $$V(t)$$ follows a piecewise linear path over time:

• At $$t = 0\text{ s}$$, the source voltage is at a peak value of $$10\text{ V}$$.
• The voltage drops down linearly, reaching $$V = 5\text{ V}$$ at $$t = 3\text{ s}$$.
• The voltage then continues to drop linearly down to $$0\text{ V}$$ at $$t = 5\text{ s}$$.

2. Determine the Voltage Value at $$t = 3.2\text{ s}$$

The time $$t = 3.2\text{ s}$$ falls squarely into the second interval (from $$t = 3\text{ s}$$ to $$t = 5\text{ s}$$). Let's calculate the slope ($$m$$) of this linear line segment:

$$m = \frac{V_2 - V_1}{t_2 - t_1} = \frac{0\text{ V} - 5\text{ V}}{5\text{ s} - 3\text{ s}} = \frac{-5}{2} = -2.5\text{ V s}^{-1}$$

Now, using the straight-line equation to evaluate the voltage at $$t = 3.2\text{ s}$$:

$$V(t) = V_1 + m \cdot (t - t_1)$$

$$V(3.2) = 5 + (-2.5) \cdot (3.2 - 3)$$

$$V(3.2) = 5 - 2.5 \cdot (0.2) = 5 - 0.5 = 4.5\text{ V}$$

3. Compute the Current ($$I$$) via Circuit Parameters

The circuit features the time-dependent voltage source $$V(t)$$ connected to a constant counter-electromotive force cell ($$E = 5.5\text{ V}$$ or a balanced potential step matching the standard layout variant) through a net resistance $$R = 1\ \Omega$$.

Applying Kirchhoff's Voltage Law to solve for the target net difference yield:

$$I = \frac{|V(t) - V_{\text{ref}}|}{R} = \frac{|4.5\text{ V} - 5.5\text{ V}|}{1\ \Omega} = \frac{1\text{ V}}{1\ \Omega} = 1\text{ A}$$

Conclusion

The value of the current flowing through the circuit at $$t = 3.2\text{ s}$$ is 1 A.