Question

Five cells, each of emf \((E)\) and internal resistance \((r)\), are connected in series. If out of these five cells, one of the cells is connected with opposite polarity, the equivalent emf and internal resistance of the combination, respectively, will be

• A. \(5E,\;5r\)
• B. \(4E,\;4r\)
• C. \(3E,\;5r\)
• D. \(3E,\;3r\)

Hint:

For cells in series: \[ E_{\text{eq}}=\sum E \] (take opposite polarity as negative) \[ r_{\text{eq}}=\sum r \] (internal resistances always add, irrespective of polarity).

Answer 1

The Correct Option is C

Concept: When cells are connected in series: \[ E_{\text{eq}}=\text{algebraic sum of emfs} \] and \[ r_{\text{eq}}=\text{sum of internal resistances} \] A cell connected in the opposite direction contributes negative emf but its internal resistance still adds positively.

Step 1:Calculate the equivalent emf. Out of five cells, \[ 4 \text{ cells contribute } +E \] and \[ 1 \text{ cell contributes } -E \] Therefore, \[ E_{\text{eq}} = 4E-E \] \[ E_{\text{eq}} = 3E \]

Step 2: Calculate the equivalent internal resistance. Internal resistances are always added in series. \[ r_{\text{eq}} = r+r+r+r+r \] \[ r_{\text{eq}} = 5r \]

Step 3: State the answer. \[ { E_{\text{eq}}=3E, \qquad r_{\text{eq}}=5r } \] Hence, the correct option is \[ {(C)} \]