The current in a circuit with an external resistance of `3.75Omega is 2.5A`. When a resistance of `1 Omega` is introduced into the circuit, the current becomes `0.4 A`. The emf of the power source is

Options

A. `1V`
B. `2V`
C. `3V`
D. `4V`

Explanation

To find the EMF of the power source in the given circuit, we can follow these steps:

### Step 1: Write down the initial conditions and equations
We are given:
- External resistance \( R = 3.75 \, \Omega \)
- Current \( I_1 = 2.5 \, A \)

Using Ohm's law, we can express the EMF \( E \) in terms of the internal resistance \( r \):
\[ I_1 = \frac{E}{R + r} \]
Substituting the known values:
\[ 2.5 = \frac{E}{3.75 + r} \quad \text{(Equation 1)} \]

### Step 2: Write down the conditions after adding a resistance
When a resistance of \( 1 \, \Omega \) is added:
- New total resistance \( R' = 3.75 + 1 = 4.75 \, \Omega \)
- New current \( I_2 = 0.4 \, A \)

Using Ohm's law again:
\[ I_2 = \frac{E}{R' + r} \implies 0.4 = \frac{E}{4.75 + r} \quad \text{(Equation 2)} \]

### Step 3: Solve Equation 1 for \( E \)
From Equation 1:
\[ E = 2.5(3.75 + r) \]
Expanding this gives:
\[ E = 9.375 + 2.5r \quad \text{(Equation 3)} \]

### Step 4: Solve Equation 2 for \( E \)
From Equation 2:
\[ E = 0.4(4.75 + r) \]
Expanding this gives:
\[ E = 1.9 + 0.4r \quad \text{(Equation 4)} \]

### Step 5: Set Equations 3 and 4 equal to each other
Since both equations represent \( E \), we can set them equal:
\[ 9.375 + 2.5r = 1.9 + 0.4r \]

### Step 6: Solve for \( r \)
Rearranging gives:
\[ 9.375 - 1.9 = 0.4r - 2.5r \]
\[ 7.475 = -2.1r \]
\[ r = -\frac{7.475}{2.1} \approx -3.55 \, \Omega \]

### Step 7: Substitute \( r \) back into either equation to find \( E \)
Using Equation 3:
\[ E = 9.375 + 2.5(-3.55) \]
Calculating this gives:
\[ E = 9.375 - 8.875 = 0.5 \, V \]

### Final Answer
The EMF of the power source is approximately:
\[ E \approx 0.5 \, V \]