question_answer

The internal resistance of cell of emf 2 V is\[0.1\,\,\Omega \]. It is connected to a resistance of \[3.9\,\,\Omega \]. The voltage across the cell is :

A)  2.71 V         

B)  1.95 V

C)  1.68 V        

D)  0.52 V

Correct Answer: B

Solution :

Key Idea: The potential difference of a cell is less than the emf of the cell. When we connect the plates of a cell by a wire, an electric current flows in the wire from positive plate towards the negative plate. From law of conservation of energy. Work done \[={{W}_{(external\text{ }circuit)}}\]                  \[+{{W}_{(internal\text{ }circuit)}}\] or            \[E\,i\,\,t=V\,\,i\,\,t+{{i}^{2}}r\,t\]                 \[\Rightarrow \] \[V=E-ir\]                 \[\Rightarrow \] \[V=E-\left( \frac{E}{R+r} \right)\,\,r\] Putting the numerical values, we have                 \[V=2-\left[ \frac{2}{3.9+0.1}\times 0.1 \right]\]                 \[=2-\frac{0.2}{4}\] \[=2-0.05=1.95\,\,V\]