A closed circuit consists of a source of constant `emf E` and a choke coil of inductance `L` connected in series.The active resistance of the whole circuit is equal to `R`.It is in steady state.At the moment `L=0` the choke coil inductance was decreased abruptly `4` times.The current in the circuit as a function of time `t` is `E/R[1-xc^(-4tR//L)]`.Find out value of `x`.

Explanation

To solve the problem, we need to analyze the circuit consisting of a constant EMF source \( E \), a choke coil with inductance \( L \), and a resistance \( R \). The choke coil's inductance is abruptly decreased to \( \frac{L}{4} \). We are given the expression for the current in the circuit as a function of time \( t \):

\[ I(t) = \frac{E}{R} \left( 1 - x e^{-\frac{4Rt}{L}} \right) \]

We need to find the value of \( x \).

### Step-by-Step Solution:

1. **Write the Circuit Equation**:
The voltage across the inductor and the resistor must equal the EMF. According to Kirchhoff's voltage law, we can write:
\[ E = I R + L \frac{dI}{dt} \]

2. **Substituting Inductance**:
Since the inductance is decreased to \( \frac{L}{4} \), we substitute this into the equation:
\[ E = I R + \frac{L}{4} \frac{dI}{dt} \]

3. **Rearranging the Equation**:
Rearranging gives us:
\[ \frac{L}{4} \frac{dI}{dt} = E - I R \]

4. **Separating Variables**:
We can express this as:
\[ \frac{dI}{E - IR} = \frac{4}{L} dt \]

5. **Integrating Both Sides**:
Integrate both sides. The left side can be integrated using the natural logarithm:
\[ \int \frac{dI}{E - IR} = -\frac{1}{R} \ln |E - IR| + C_1 \]
The right side integrates to:
\[ \int \frac{4}{L} dt = \frac{4}{L} t + C_2 \]

6. **Combining Constants**:
Setting the constants equal gives:
\[ -\frac{1}{R} \ln |E - IR| = \frac{4}{L} t + C \]

7. **Solving for Current**:
Exponentiating both sides leads to:
\[ E - IR = K e^{-\frac{4}{L} t} \]
where \( K \) is a constant determined by initial conditions.

8. **Finding the Current**:
Rearranging gives:
\[ I = \frac{E}{R} - \frac{K}{R} e^{-\frac{4}{L} t} \]
Let \( K = 3E \) (based on the steady state where \( I \) approaches \( \frac{E}{R} \)):
\[ I = \frac{E}{R} - \frac{3E}{R} e^{-\frac{4}{L} t} \]
Therefore:
\[ I = \frac{E}{R} \left( 1 - 3 e^{-\frac{4}{L} t} \right) \]

9. **Comparing with Given Expression**:
Now we compare this with the given expression:
\[ I(t) = \frac{E}{R} \left( 1 - x e^{-\frac{4Rt}{L}} \right) \]
From this comparison, we see that \( x = 3 \).

### Final Answer:
The value of \( x \) is \( 3 \).