Question:
When an external resistor of resistance \(18\,\Omega\) is connected to a cell, the current drawn from the cell is \(I\). If another \(18\,\Omega\) resistor is connected parallel to the first resistor, the current drawn from the cell increases by \(90\%\), then the internal resistance of the cell is
When an external resistor of resistance \(18\,\Omega\) is connected to a cell, the current drawn from the cell is \(I\). If another \(18\,\Omega\) resistor is connected parallel to the first resistor, the current drawn from the cell increases by \(90\%\), then the internal resistance of the cell is
Show Hint
For a cell,
\[
I=\frac{E}{R+r}.
\]
Whenever the external resistance changes, write the current expression before and after the change, then use the given percentage increase/decrease to determine the internal resistance.
Updated On: Jul 29, 2026
- \(2\,\Omega\)
- \(0.5\,\Omega\)
- \(1.5\,\Omega\)
- \(1\,\Omega\)
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The Correct Option is D
Solution and Explanation
Concept:
Current supplied by a cell is
\[
I=\frac{E}{R+r},
\]
where
\[
E=\text{emf of the cell},
\qquad
R=\text{external resistance},
\qquad
r=\text{internal resistance}.
\]
Step 1: Find the initial current. Initially, \[ R=18\,\Omega. \] Hence, \[ I_1=\frac{E}{18+r}. \]
Step 2: Find the new current when another \(18\,\Omega\) resistor is connected in parallel. Equivalent resistance: \[ R' = \frac{18\times18}{18+18}. \] \[ R'=9\,\Omega. \] Therefore, \[ I_2=\frac{E}{9+r}. \]
Step 3: Use the given condition that current increases by \(90\%\). \[ I_2=1.9I_1. \] Substituting, \[ \frac{E}{9+r} = 1.9 \left( \frac{E}{18+r} \right). \] Cancelling \(E\), \[ 18+r = 1.9(9+r). \] \[ 18+r = 17.1+1.9r. \] \[ 0.9 = 0.9r. \] \[ r=1\,\Omega. \] Therefore, \[ \boxed{r=1\,\Omega} \] \[ \boxed{\text{Answer = (D)}} \]
Step 1: Find the initial current. Initially, \[ R=18\,\Omega. \] Hence, \[ I_1=\frac{E}{18+r}. \]
Step 2: Find the new current when another \(18\,\Omega\) resistor is connected in parallel. Equivalent resistance: \[ R' = \frac{18\times18}{18+18}. \] \[ R'=9\,\Omega. \] Therefore, \[ I_2=\frac{E}{9+r}. \]
Step 3: Use the given condition that current increases by \(90\%\). \[ I_2=1.9I_1. \] Substituting, \[ \frac{E}{9+r} = 1.9 \left( \frac{E}{18+r} \right). \] Cancelling \(E\), \[ 18+r = 1.9(9+r). \] \[ 18+r = 17.1+1.9r. \] \[ 0.9 = 0.9r. \] \[ r=1\,\Omega. \] Therefore, \[ \boxed{r=1\,\Omega} \] \[ \boxed{\text{Answer = (D)}} \]
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