JEE MAIN - Physics (2018 (Offline) - No. 8)
Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of 10 $$\Omega $$. The
internal resistances of the two batteries are 1 $$\Omega $$ and 2 $$\Omega $$ respectively. The voltage across the load lies between :
11.7 V and 11.8 V
11.6 V and 11.7 V
11.5 V and 11.6 V
11.4 V and 11.5 V
Explanation
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Let potential at S, T, R = V and potential at P, Q, U, = 0
Using kirchhoff's law at P :
Current at P is ,
$${{V - 12} \over 1} + {{V - 13} \over 2} + {{V - 0} \over {10}} = 0$$
$$ \Rightarrow $$$$\,\,\,$$ $${V \over 1}$$ + $${V \over 2}$$ + $${V \over 10}$$ = 12 + $${{13} \over 2}$$
$$ \Rightarrow $$$$\,\,\,$$ $${{10V + 5V + V} \over {10}}$$ = $${{37} \over 2}$$
$$ \Rightarrow $$$$\,\,\,$$ $${{16V} \over {10}}$$ = $${{37} \over 2}$$
$$ \Rightarrow $$$$\,\,\,$$ V = 11.56 Volt.
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