A) 1 ohm
B) 0.2 ohm
C) 2.5 ohm
D) 0.4 ohm
Correct Answer: D
Solution :
The potential difference in open circuit for the cell is 2.2 volts, it means the emf of the battery should be 2.2 volts. Now a 4 ohm resistor is connected between its two electrodes, then the potential difference across the battery becomes 2 volts \[V=\varepsilon -ir\Rightarrow 2=2.2-ir\] \[\Rightarrow \] \[ir=0.2\,V\] ?(i) But \[i=\frac{e}{r+R}=\frac{2.2}{r+4}\] ?(ii) Putting the value of \[i\] in (i), we get r = 0.4 ohm.You need to login to perform this action.
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