n identical cells each of e.m.f. E and internal resistance r are connected in series. An external resistance R is connected in series to this combination. The current through R is

Options

A. `(nE)/(R + nr)`
B. `(nE)/(nR + r)`
C. `E/(R + nr)`
D. `(nE)/(R + r)`

Answer

To solve the problem, we need to determine the current flowing through the external resistance \( R \) when \( n \) identical cells, each with an electromotive force (e.m.f.) \( E \) and internal resistance \( r \), are connected in series with the external resistance \( R \).

### Step-by-Step Solution:

1. **Identify the Total e.m.f. in the Circuit:**
When \( n \) identical cells are connected in series, the total e.m.f. \( E_{\text{total}} \) of the combination is given by:
\[ E_{\text{total}} = nE \]

2. **Identify the Total Internal Resistance:**
The internal resistance of each cell is \( r \). Since they are connected in series, the total internal resistance \( r_{\text{total}} \) of the combination is:
\[ r_{\text{total}} = nr \]

3. **Set Up the Circuit Equation:**
According to Kirchhoff's voltage law, the sum of the potential differences (voltage) in a closed circuit must equal zero. For our circuit, we have:
\[ E_{\text{total}} = I(R + r_{\text{total}}) \]
Substituting the values we found:
\[ nE = I(R + nr) \]

4. **Solve for the Current \( I \):**
Rearranging the equation to solve for the current \( I \):
\[ I = \frac{nE}{R + nr} \]

5. **Final Expression for Current:**
Thus, the current through the external resistance \( R \) is:
\[ I = \frac{nE}{R + nr} \]