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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
1031890_16c25dbdfeea4a6aa6695fac135a24d2.png


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

The correct option is C $$\dfrac { E\left( R+4r \right) }{ R+3r }$$

$$\hbox{So, circuit current}, I$$

$$=\dfrac{E}{R+3r}$$

$$\hbox{P.D. across C}$$

$$=Ir+E$$

$$=\dfrac{Er}{R+3r}+E$$

$$=E\left(\dfrac{R+4r}{R+3r} \right )$$


1007028_1031890_ans_1d824c794af84da5850f7cf34e993af2.PNG

Physics

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