Question

# Three cell, $$(A,B$$ & $$C)$$ each of emf $$E$$ and internal resistance $$(r)$$ are connected in series as shown in figure. The potential difference across the cell $$'C'$$ is

A
ErR
B
E(R+r)R+3r
C
E(R+4r)R+3r
D
E(R+3r)R+4r

Solution

## The correct option is C $$\dfrac { E\left( R+4r \right) }{ R+3r }$$$$\hbox{So, circuit current}, I$$ $$=\dfrac{E}{R+3r}$$ $$\hbox{P.D. across C}$$ $$=Ir+E$$ $$=\dfrac{Er}{R+3r}+E$$ $$=E\left(\dfrac{R+4r}{R+3r} \right )$$Physics

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