A battery having 12V emf and internal resistance `3Omega` is connected to a resistor. If the current in the circuit is 1A, then the resistance of resistor and lost voltage of the battery when circuit is closed will be

Options

A. `7Omega,7V`
B. `8Omega,8V`
C. `9Omega,9V`
D. `9Omega,10V`

Explanation

To solve the problem step by step, we will find the external resistance and the lost voltage of the battery when the circuit is closed.

### Step 1: Understand the Circuit
We have a battery with an EMF (Electromotive Force) of 12V and an internal resistance of 3Ω. The battery is connected to an external resistor (let's call it R) and the current flowing through the circuit is given as 1A.

### Step 2: Apply Ohm's Law
According to Ohm's law, the total voltage in the circuit is equal to the sum of the voltage drops across the internal resistance and the external resistance. The formula can be expressed as:
\[ I = \frac{E}{R + r} \]
Where:
- \( I \) = current (1A)
- \( E \) = EMF of the battery (12V)
- \( R \) = external resistance (unknown)
- \( r \) = internal resistance (3Ω)

### Step 3: Substitute Known Values
Substituting the known values into the equation:
\[ 1 = \frac{12}{R + 3} \]

### Step 4: Solve for R
Cross-multiplying gives:
\[ R + 3 = 12 \]
Now, isolate R:
\[ R = 12 - 3 \]
\[ R = 9Ω \]

### Step 5: Calculate the Lost Voltage
The lost voltage (or terminal voltage) of the battery can be calculated using the formula:
\[ V = E - I \cdot r \]
Where:
- \( V \) = terminal voltage
- \( E \) = EMF of the battery (12V)
- \( I \) = current (1A)
- \( r \) = internal resistance (3Ω)

Substituting the known values:
\[ V = 12 - (1 \cdot 3) \]
\[ V = 12 - 3 \]
\[ V = 9V \]

### Final Results
- The resistance of the external resistor \( R \) is **9Ω**.
- The lost voltage (terminal voltage) of the battery is **9V**.