A) 3.9 V
B) 2 V
C) 0.1 V
D) 3.8 V
Correct Answer: A
Solution :
Emf of a cell \[(E)\,=\,4\,V\] Internal resistance of a cell \[(r)=0.1\,\Omega \] External resistance \[(R)=3.9\,\Omega \] The potential drop across the cell \[V=E-I\cdot r\] ...(i) Now, the total resistance of the circuit \[R'=r+R\] \[R'=0.1+3.9\] \[\Rightarrow \] \[R'=4.0\,\Omega \] Hence, current in the circuit is \[I=\frac{E}{R'}\] \[\Rightarrow \] \[I=\frac{4}{4}\,=1\,A\] Now, from Eq. (i) \[V=4\,-1\,\,\times 0.1\] \[V=4-0.1\] \[V=3.9\] volt
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