JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 20)
The series combination of two batteries, both
of the same emf 10 V, but different internal
resistance of 20$$\Omega $$ and 5$$\Omega $$, is connected to the
parallel combination of two resistors 30$$\Omega $$ and
R $$\Omega $$. The voltage difference across the battery
of internal resistance 20$$\Omega $$ is zero, the value of
R (in $$\Omega $$) is : _______
Answer
30
Explanation
_8th_January_Evening_Slot_en_20_1.png)
10 – I × 20 = 0
$$ \Rightarrow $$ I = 0.5A
Also,
I = $${{10 + 10} \over {20 + 5 + {{30R} \over {30 + R}}}}$$
$$ \Rightarrow $$ $${1 \over 2}$$ = $${{20} \over {25 + {{30R} \over {30 + R}}}}$$
$$ \Rightarrow $$ 40 = 25 + $${{{30R} \over {30 + R}}}$$
$$ \Rightarrow $$ 15 = $${{{30R} \over {30 + R}}}$$
$$ \Rightarrow $$ 30 + R = 2R
$$ \Rightarrow $$ R = 30 $$\Omega $$
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