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

Explanation

To solve the problem, we will use the formula for the time constant of an inductive circuit, 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 given time constants.

1. The first time constant is given as \(2 \times 10^{-3} \, s\):
\[ \frac{L}{R} = 2 \times 10^{-3} \tag{1} \]

2. The second time constant, when a \(90 \, \Omega\) resistor is added in series, is \(0.5 \times 10^{-3} \, s\):
\[ \frac{L}{R + 90} = 0.5 \times 10^{-3} \tag{2} \]

### Step 2: Solve the equations.

From equation (1):
\[ L = 2 \times 10^{-3} R \tag{3} \]

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

### Step 3: Cross-multiply to eliminate the fraction.

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

### Step 4: Distribute on the right side.

\[ 2 \times 10^{-3} R = 0.5 \times 10^{-3} R + 0.5 \times 10^{-3} \times 90 \]

Calculating \(0.5 \times 10^{-3} \times 90\):
\[ 0.5 \times 90 = 45 \quad \Rightarrow \quad 0.5 \times 10^{-3} \times 90 = 45 \times 10^{-3} \]

So we have:
\[ 2 \times 10^{-3} R = 0.5 \times 10^{-3} R + 45 \times 10^{-3} \]

### Step 5: Rearranging the equation.

Subtract \(0.5 \times 10^{-3} R\) from both sides:
\[ 2 \times 10^{-3} R - 0.5 \times 10^{-3} R = 45 \times 10^{-3} \]

This simplifies to:
\[ 1.5 \times 10^{-3} R = 45 \times 10^{-3} \]

### Step 6: 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 7: Substitute \(R\) back to find \(L\).

Using equation (3):
\[ L = 2 \times 10^{-3} \times 30 = 60 \times 10^{-3} \, H = 60 \, mH \]

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