A) 1.95 V
B) 12.7 V
C) 1.5 V
D) 6.71 V
Correct Answer: A
Solution :
When current is drawn from cell, its emf (E) reduces and potential drop (V) is given by \[V=E-Ir\] ... (i) From Ohms law \[I=\frac{E}{R}\] ... (ii) \[R=3.9\,\,\Omega +0.1\,\,\Omega =4\,\,\Omega \], \[E=2\,V,\,\,\,r=0.1\,\,\Omega \] \[\because \] \[I=\frac{2}{4}=0.5\] ... (iii) For Eqs. (i), (ii) and (iii), we get \[V=2-0.5\times 0.1\] \[\Rightarrow \] V = 1.95 voltYou need to login to perform this action.
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