The time constant of a circuit is 10 sec , when a resistance of `10 Omega` is connected in series in a prevouse circuit then time constant becomes 2 second, then the self inductance of the circuit is :-

Options

A. 2.5 H
B. 5H
C. 15H
D. 25H

Answer

To find the self-inductance of the circuit, we can follow these steps:

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

### Step 2: Set Up the Initial Condition
Initially, the time constant is given as 10 seconds. Let the resistance in the circuit be \(R\). Therefore, we can write:
\[ \tau = \frac{L}{R} = 10 \quad \text{(1)} \]
From this, we can express \(L\) in terms of \(R\):
\[ L = 10R \quad \text{(2)} \]

### Step 3: Set Up the Condition with Additional Resistance
When a resistance of \(10 \, \Omega\) is added in series, the new equivalent resistance becomes \(R + 10\). The new time constant is given as 2 seconds:
\[ \tau' = \frac{L}{R + 10} = 2 \quad \text{(3)} \]

### Step 4: Express \(L\) from the New Condition
From equation (3), we can express \(L\) as:
\[ L = 2(R + 10) \quad \text{(4)} \]

### Step 5: Equate the Two Expressions for \(L\)
Now we have two expressions for \(L\) from equations (2) and (4):
1. \(L = 10R\)
2. \(L = 2(R + 10)\)

Setting these equal to each other:
\[ 10R = 2(R + 10) \]

### Step 6: Solve for \(R\)
Expanding the right side:
\[ 10R = 2R + 20 \]
Subtracting \(2R\) from both sides:
\[ 8R = 20 \]
Dividing both sides by 8:
\[ R = \frac{20}{8} = 2.5 \, \Omega \]

### Step 7: Substitute \(R\) Back to Find \(L\)
Now substituting \(R\) back into equation (2) to find \(L\):
\[ L = 10R = 10 \times 2.5 = 25 \, H \]

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

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