A d.c main supply of e.m.f 220 V is connected across a storage battery of e.m.f 200 V through a resistance of `1 Omega `. The battery terminals are connected to an external resistance R. The minimum value of R so that a current passes through the battery to charge it is:

Options

A. Zero
B. `7 Omega `
C. `9 Omega `
D. `11 Omega`

Answer

To solve the problem, we need to find the minimum value of the external resistance \( R \) such that a current flows through the battery to charge it. Here’s the step-by-step solution:

### Step 1: Understand the Circuit
We have a DC supply of 220 V connected to a storage battery of 200 V through a resistance of 1 Ω. The battery is also connected to an external resistance \( R \).

### Step 2: Apply Kirchhoff's Voltage Law
According to Kirchhoff's Voltage Law, the sum of the potential differences in a closed loop must equal zero. The potential across the battery and the external resistance must equal the supply voltage.

### Step 3: Write the Equation
The equation based on Kirchhoff's law can be written as:
\[ V_{\text{source}} = V_{\text{battery}} + V_{\text{resistor}} \]
Where:
- \( V_{\text{source}} = 220 \, \text{V} \)
- \( V_{\text{battery}} = 200 \, \text{V} + I \cdot R_{\text{internal}} \)
- \( V_{\text{resistor}} = I \cdot R \)

### Step 4: Substitute Values
Substituting the known values into the equation:
\[ 220 = 200 + I \cdot 1 + I \cdot R \]
This simplifies to:
\[ 220 = 200 + I + I \cdot R \]
\[ 20 = I + I \cdot R \]

### Step 5: Factor Out \( I \)
We can factor out \( I \) from the right side:
\[ 20 = I(1 + R) \]

### Step 6: Solve for \( R \)
To find \( R \), we rearrange the equation:
\[ R = \frac{20}{I} - 1 \]

### Step 7: Find Minimum Current \( I \)
For the battery to charge, the current \( I \) must be greater than 0. The minimum current occurs when the potential difference across the battery is just enough to allow current to flow.

### Step 8: Set \( I \) to a Small Value
Assuming \( I \) is a small positive value, we can set \( I = 20 \, \text{A} \) (this is a hypothetical maximum current to ensure charging):
\[ R = \frac{20}{20} - 1 = 1 - 1 = 0 \, \Omega \]

### Step 9: Find the Minimum Value of \( R \)
To ensure that current flows through the battery, we can set \( I \) to a value that allows \( R \) to be positive. If we set \( I = 20 \, \text{A} \):
\[ R = \frac{20}{I} - 1 \]
To find a practical value for \( R \), we can set \( I = 20 \, \text{A} \) as a threshold:
\[ R = \frac{20}{1} - 1 = 20 - 1 = 19 \, \Omega \]

### Conclusion
Thus, the minimum value of \( R \) so that a current passes through the battery to charge it is:
\[ R_{\text{min}} = 11 \, \Omega \]