JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 16)
Two resistors 400$$\Omega $$ and 800$$\Omega $$ are connected in series across a 6 V battery. The potential difference measured by a voltmeter of 10 k$$\Omega $$ across 400 $$\Omega $$ resistor is close to :
2.05 V
1.95 V
2 V
1.8 V
Explanation
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$$i = {6 \over {800 + {{400 \times 10000} \over {400 + 10000}}}}$$
$$i = {6 \over {800 + {{40000} \over {104}}}}$$
$$i = {6 \over {800 + 384.61}} = {6 \over {1184.61}} = 0.00506$$
V400 = i1 $$ \times $$ 400
= $$\left( {{{{{10}^4}} \over {400 + {{10}^4}}}} \right)i \times 400$$
= $$\left( {{{{{10}^4}} \over {400 + {{10}^4}}}} \right)\left( {0.00506} \right) \times 400$$
= 1.95 V
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