Series RL Circuit Current Calculation at 0.1 sec

This question involves calculating the current flowing through a series RL circuit after a DC voltage is applied. We are given the values for inductance, resistance, the applied voltage, and the specific time at which we need to find the current.

Understanding the Circuit Parameters

• Type of Circuit: Series RL Circuit
• Inductance (L): 1 Henry (H)
• Resistance (R): 10 Ohms ($\Omega$)
• Applied Voltage (V): 100 Volts (V) DC
• Time (t): 0.1 seconds (sec)

Formula for Transient Current in DC RL Circuit

When a DC voltage is applied to a series RL circuit at time $t=0$, the current does not reach its final steady-state value instantaneously. It grows exponentially over time. The formula describing the current $I(t)$ at any time $t$ is:

$$I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right)$$

Where:

• $I(t)$ is the current at time $t$.
• $V$ is the applied DC voltage.
• $R$ is the resistance.
• $L$ is the inductance.
• $e$ is the base of the natural logarithm (approximately 2.71828).
• $t$ is the time in seconds.

Calculating the Time Constant

The term $\frac{L}{R}$ in the exponent is known as the time constant ($\tau$) of the RL circuit. It represents the time required for the current to reach approximately 63.2% of its final steady-state value.

Calculating the time constant for this circuit:

$$\tau = \frac{L}{R}$$

Substituting the given values:

$$\tau = \frac{1 \text{ H}}{10 \text{ } \Omega} = 0.1 \text{ sec}$$

Calculating Current at t = 0.1 sec

Now, we substitute the circuit parameters and the time $t=0.1$ sec into the current formula:

$$I(0.1) = \frac{100 \text{ V}}{10 \text{ } \Omega} \left(1 - e^{-\frac{10 \text{ } \Omega}{1 \text{ H}} \times 0.1 \text{ sec}}\right)$$

Simplify the expression:

$$I(0.1) = 10 \text{ A} \left(1 - e^{-(10 \times 0.1)}\right)$$

$$I(0.1) = 10 \text{ A} \left(1 - e^{-1}\right)$$

We know that $e^{-1}$ is approximately 0.367879.

$$I(0.1) = 10 \text{ A} (1 - 0.367879)$$

$$I(0.1) = 10 \text{ A} (0.632121)$$

$$I(0.1) \approx 6.32121 \text{ A}$$

Conclusion

The value of the current in the series RL circuit at 0.1 seconds is approximately 6.32 A. This matches one of the provided options.