The time constant of an inductance coil is `2 xx 10^(-3) s`. When a `90 Omega` resistance is joined in series, the same constant becomes `0.5 xx 10^(-3) s`. The inductance and resistance of the coil are

Options

A. 30mH,`30Omega`
B. 30mH,`60Omega`
C. 60mH,`30Omega`
D. 60mH,`60Omega`

Explanation

To solve the problem, we need to use the formula for the time constant of an inductor, which is given by:

\[ \tau = \frac{L}{R} \]

where:
- \(\tau\) is the time constant,
- \(L\) is the inductance,
- \(R\) is the resistance.

### Step 1: Set up the equations based on the time constants.

1. **Initial Time Constant**:
Given that the initial time constant \(\tau_1 = 2 \times 10^{-3} \, \text{s}\), we can write:
\[ \tau_1 = \frac{L}{R} \quad \text{(1)} \]

2. **New Time Constant with Added Resistance**:
When a \(90 \, \Omega\) resistor is added in series, the new time constant \(\tau_2 = 0.5 \times 10^{-3} \, \text{s}\). Thus, we can express this as:
\[ \tau_2 = \frac{L}{R + 90} \quad \text{(2)} \]

### Step 2: Substitute and solve the equations.

From equation (1), we can express \(L\) in terms of \(R\):
\[ L = \tau_1 \cdot R = 2 \times 10^{-3} \cdot R \quad \text{(3)} \]

Now substitute equation (3) into equation (2):
\[ 0.5 \times 10^{-3} = \frac{2 \times 10^{-3} \cdot R}{R + 90} \]

### Step 3: Cross-multiply and simplify.

Cross-multiplying gives:
\[ 0.5 \times 10^{-3} (R + 90) = 2 \times 10^{-3} R \]

Expanding this:
\[ 0.5 \times 10^{-3} R + 45 \times 10^{-3} = 2 \times 10^{-3} R \]

### Step 4: Rearrange the equation.

Rearranging the terms gives:
\[ 45 \times 10^{-3} = 2 \times 10^{-3} R - 0.5 \times 10^{-3} R \]
\[ 45 \times 10^{-3} = (2 - 0.5) \times 10^{-3} R \]
\[ 45 \times 10^{-3} = 1.5 \times 10^{-3} R \]

### Step 5: Solve for \(R\).

Dividing both sides by \(1.5 \times 10^{-3}\):
\[ R = \frac{45 \times 10^{-3}}{1.5 \times 10^{-3}} = 30 \, \Omega \]

### Step 6: Substitute \(R\) back to find \(L\).

Now substituting \(R = 30 \, \Omega\) back into equation (3) to find \(L\):
\[ L = 2 \times 10^{-3} \cdot 30 = 60 \times 10^{-3} \, \text{H} = 60 \, \text{mH} \]

### Final Answer:
- The resistance \(R\) is \(30 \, \Omega\).
- The inductance \(L\) is \(60 \, \text{mH}\).