KCET · Physics · Electrostatics
Four identical cells of emf \( E \) and internal resistance \( r \) are to be connected in series. Suppose if one of the cell is connected wrongly, the equivalent emf and effective internal resistance of the combination is
- A \( 4 \mathrm{E} \) and \( 4 \mathrm{r} \)
- B \( 4 \mathrm{E} \) and \( 2 r \)
- C \( 2 \mathrm{E} \) and \( 4 \mathrm{r} \)
- D \( 2 \mathrm{E} \) and \( 2 r \)
Answer & Solution
Correct Answer
(C) \( 2 \mathrm{E} \) and \( 4 \mathrm{r} \)
Step-by-step Solution
Detailed explanation
When all the cells are connected correctly, then
Equivalent emf = 4E and effective resistance \( =4 r \)
If one of the cell is connected wrongly then, effective internal resistance will remain unchanged while, equivalent emf is
\( E^{\prime}=E(n-2 m) \)
where \( n \) is total number of cells; \( m \) is number of wrongly connected cells.
Therefore,
\( E=E(4-2 \times 1)=E(4-2)=2 E \)
\( \Rightarrow \) equivalent emf \( =2 \mathrm{E} \) and effective internal resistance \( =4 \mathrm{r} \)
Equivalent emf = 4E and effective resistance \( =4 r \)
If one of the cell is connected wrongly then, effective internal resistance will remain unchanged while, equivalent emf is
\( E^{\prime}=E(n-2 m) \)
where \( n \) is total number of cells; \( m \) is number of wrongly connected cells.
Therefore,
\( E=E(4-2 \times 1)=E(4-2)=2 E \)
\( \Rightarrow \) equivalent emf \( =2 \mathrm{E} \) and effective internal resistance \( =4 \mathrm{r} \)
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