The time constunt of an `L-R` circuit is `10 s`. When a resistance of `10 ohm` is connected in series in a previous circuit, then the time constant becomes. `2s`. The self-inductance (in henry) of the circuit is

Answer

To solve the problem, we need to find the self-inductance (L) of the circuit given the time constants before and after a resistance is added. Let's break it down step by step.

### Step 1: Understand the Time Constant Formula
The time constant (τ) of an LR circuit is given by the formula:
\[ \tau = \frac{L}{R} \]
where \(L\) is the self-inductance in henries and \(R\) is the resistance in ohms.

### Step 2: Set Up the Equations
From the problem, we know:
1. The initial time constant (τ₁) is 10 seconds when the resistance is \(R\).
\[ \tau_1 = \frac{L}{R} = 10 \quad \text{(1)} \]
2. After adding a resistance of 10 ohms, the new time constant (τ₂) is 2 seconds.
\[ \tau_2 = \frac{L}{R + 10} = 2 \quad \text{(2)} \]

### Step 3: Express L in Terms of R
From equation (1):
\[ L = 10R \quad \text{(3)} \]

### Step 4: Substitute L in the Second Equation
Now, substitute equation (3) into equation (2):
\[ 2 = \frac{10R}{R + 10} \]

### Step 5: Cross-Multiply to Solve for R
Cross-multiplying gives:
\[ 2(R + 10) = 10R \]
Expanding this:
\[ 2R + 20 = 10R \]
Rearranging gives:
\[ 10R - 2R = 20 \]
\[ 8R = 20 \]
\[ R = \frac{20}{8} = 2.5 \, \text{ohms} \]

### Step 6: Substitute R Back to Find L
Now substitute \(R\) back into equation (3) to find \(L\):
\[ L = 10R = 10 \times 2.5 = 25 \, \text{henries} \]

### Final Answer
The self-inductance of the circuit is:
\[ \boxed{25 \, \text{henries}} \]