Question

Five cells each of emf \(E\) and internal resistance \(r\) send the same amount of current through an external resistance \(R\) whether the cells are connected in parallel or in series. Then the ratio \(\frac{R}{r}\) is:

• A. \(2\)
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{5}\)
• D. \(1\)

Hint:

For the same current in both series and parallel configurations, the external resistance must equal internal resistance.

Answer 1

The Correct Option is D

Step 1: {Current in series combination}
\[ I = \frac{nE}{nr + R} = \frac{5E}{5r + R} \] Step 2: {Current in parallel combination}
\[ I' = \frac{E}{\frac{r}{n} + R} = \frac{5E}{r + 5R} \] Since \(I = I'\), equating both expressions: \[ \frac{5E}{5r + R} = \frac{5E}{r + 5R} \] Solving for \(R\) and \(r\), \[ 5r + R = r + 5R \] \[ 4r = 4R \Rightarrow R = r \] Thus, the ratio \(\frac{R}{r} = 1\).

Answer 2

Step 1: Case 1 – Cells connected in series
In series, the total emf = \( 5E \)
Total internal resistance = \( 5r \)
Using Ohm’s law, the total current through the external resistance \( R \) is:
\( I_s = \frac{5E}{R + 5r} \)

Step 2: Case 2 – Cells connected in parallel
In parallel, the equivalent emf remains \( E \) (since all have same emf)
The equivalent internal resistance becomes \( \frac{r}{5} \)
So the current through the same external resistance \( R \) is:
\( I_p = \frac{E}{R + \frac{r}{5}} \)

Step 3: Given condition
The current through \( R \) is the same in both cases:
So, \( I_s = I_p \)
\[ \frac{5E}{R + 5r} = \frac{E}{R + \frac{r}{5}} \]

Step 4: Cancel \( E \) from both sides and cross-multiply
\[ \frac{5}{R + 5r} = \frac{1}{R + \frac{r}{5}} \]
Cross-multiplying:
\[ 5 \left( R + \frac{r}{5} \right) = R + 5r \]
\[ 5R + r = R + 5r \]

Step 5: Simplify the equation
\[ 5R - R = 5r - r \Rightarrow 4R = 4r \Rightarrow R = r \]

Final Answer:
\( \frac{R}{r} = 1 \)