A coil of inductanece `5H` is joined to a cell of emf `6V` through a resistance `10 Omega` at time t=0. The emf across the coil at time `t=ln sqrt(2)` s is:

Options

A. 3 V
B. 1.5 V
C. 0.75 V
D. 4.5 V

Answer

To solve the problem, we need to find the emf (electromotive force) across the coil (inductor) at a specific time \( t = \ln \sqrt{2} \) seconds. The given values are:

- Inductance \( L = 5 \, H \)
- Emf of the cell \( E = 6 \, V \)
- Resistance \( R = 10 \, \Omega \)

### Step-by-Step Solution:

1. **Understand the Circuit Behavior**:
- At \( t = 0 \), the inductor behaves like an open circuit, and no current flows.
- As time progresses, the inductor allows current to flow, reaching a maximum steady state at \( t = \infty \).

2. **Current in the Inductor**:
- The current \( I(t) \) through the inductor at any time \( t \) is given by the formula:
\[ I(t) = I_0 \left(1 - e^{-\frac{R}{L} t}\right) \]
- Here, \( I_0 \) is the maximum current, which can be calculated as:
\[ I_0 = \frac{E}{R} = \frac{6 \, V}{10 \, \Omega} = 0.6 \, A \]

3. **Substituting Values**:
- Substitute \( I_0 \) into the current equation:
\[ I(t) = 0.6 \left(1 - e^{-\frac{10}{5} t}\right) = 0.6 \left(1 - e^{-2t}\right) \]

4. **Finding \( \frac{dI}{dt} \)**:
- Differentiate \( I(t) \) with respect to \( t \):
\[ \frac{dI}{dt} = 0.6 \cdot \frac{d}{dt}\left(1 - e^{-2t}\right) = 0.6 \cdot (0 + 2e^{-2t}) = 1.2 e^{-2t} \]

5. **Calculate the Voltage across the Inductor**:
- The voltage \( V_L \) across the inductor is given by:
\[ V_L = L \frac{dI}{dt} \]
- Substitute \( L \) and \( \frac{dI}{dt} \):
\[ V_L = 5 \cdot (1.2 e^{-2t}) = 6 e^{-2t} \]

6. **Evaluate at \( t = \ln \sqrt{2} \)**:
- Substitute \( t = \ln \sqrt{2} \):
\[ V_L = 6 e^{-2 \ln \sqrt{2}} = 6 e^{\ln(2^{-1})} = 6 \cdot \frac{1}{2} = 3 \, V \]

### Final Answer:
The emf across the coil at time \( t = \ln \sqrt{2} \) seconds is **3 V**.