Question

Four identical cells, each with internal resistance 1 \(\Omega\) and emf 10 V are connected in series to a resistance R and they are connected in parallel and then to an external resistance R in series. For what value of R the current in series and in parallel combination is the same?

• A. \(1 \, \Omega\)
• B. \(4 \, \Omega\)
• C. \(2 \, \Omega\)
• D. \(6 \, \Omega\)

Hint:

When balancing series and parallel circuits to achieve equal currents, equating the total resistances can simplify finding the necessary external resistance.

Answer 1

The Correct Option is A

Step 1: Calculate the total emf and resistance in series.

\[ \text{Total emf (E)} = 4 \times 10 \, \text{V} = 40 \, \text{V} \] \[ \text{Total resistance (R)} = 4 \times 1 \, \Omega + R = 4 \, \Omega + R \]

Step 2: Calculate the total emf and resistance in parallel.

For parallel, the total resistance of the batteries alone: \[ \frac{1}{R_{\text{total}}} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 4 \] \[ R_{\text{total}} = 0.25 \, \Omega \] Total resistance with external \( R \): \[ R_{\text{parallel}} = 0.25 \, \Omega + R \]

Step 3: Set the currents equal for series and parallel circuits.

\[ \frac{40}{4+R} = \frac{10}{0.25+R} \] Solving for \( R \), \[ R = 1 \, \Omega \]