Question

A coil of inductance 300mH and resistance 2Ω is connected to a source of voltage 2v. The current reaches half of its steady-state value in____seconds.

• A. 0.1
• B. 0.3
• C. 0.05
• D. 0.105

Answer 1

The correct option is A

0.1

Step 1: Given

Inductance, L=300mH=$300\times {10}^{-3}H$

Resistance, R= 2Ω

Voltage, V=2V

Step 2: Use the formula of growth of current

Formula for the growth of current in inductor, $I=I_0\left(1-e^{-\frac{Rt}{L}}\right)$

Also, $I=\frac{I_0}{2}$ because the time requires is for the half of current steady-state value.

Now,

$$\begin{gathered} \frac{I_0}{2}=I_0\left(1-e^{-\frac{Rt}{L}}\right) \\ \frac{I_0}{2}=I_0-I_0\left(e^{-\frac{Rt}{L}}\right) \\ {I}_0\left(e^{-\frac{Rt}{L}}\right)=I_0-\frac{I_0}{2} \\ {I}_0\left(e^{-\frac{Rt}{L}}\right)=\frac{I_0}{2} \\ {e}^{-\frac{Rt}{L}}=\frac{1}{2} \end{gathered}$$

taking log on both sides-

$$\begin{gathered} -\frac{Rt}{L}=-\log 2 \\ \frac{Rt}{L}=\log 2 \end{gathered}$$

Step 3: Calculate the time to reach half of its steady-state value

$t=\frac{L}{R}\log 2$

Substitute values-

$$\begin{gathered} t=\frac{300\times {10}^{-3}}{2}\times 0.693 \\ t=150\times {10}^{-3}\times 0.693 \\ t=0.10395s\approx 0.1s \end{gathered}$$

Thus, the time to reach half of its steady-state value is $0.10395s\approx 0.1s$.

Hence, Option A is correct.