JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 1)
A cell of internal resistance r drives current
through an external resistance R. The power
delivered by the cell to the external resistance
will be maximum when :-
R = 1000 r
R = r
R = 2r
R = 0.001 r
Explanation
Current i = $${E \over {r + R}}$$
Power generated in R
P = i2R
$$P = {{{E^2}R} \over {{{\left( {r + R} \right)}^2}}}$$
For maximum power $${{dP} \over {dR}} = 0$$
$${E^2}\left[ {{{{{\left( {r + R} \right)}^2} \times 1 - R \times 2(r + R)} \over {{{\left( {r + R} \right)}^4}}}} \right] = 0$$
$$ \Rightarrow $$ r = R or R = r
Power generated in R
P = i2R
$$P = {{{E^2}R} \over {{{\left( {r + R} \right)}^2}}}$$
For maximum power $${{dP} \over {dR}} = 0$$
$${E^2}\left[ {{{{{\left( {r + R} \right)}^2} \times 1 - R \times 2(r + R)} \over {{{\left( {r + R} \right)}^4}}}} \right] = 0$$
$$ \Rightarrow $$ r = R or R = r
Comments (0)
