JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 9)
A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is :
0.50 W
0.072 W
0.10 W
0.125 W
Explanation
PR = 0.5 W
$$ \Rightarrow $$ i2R = 0.5 W
iR = 2.5
$$ \Rightarrow $$ i = 0.2 A & R = 12.5 $$\Omega $$
Also, V = E – ir
$$ \Rightarrow $$ 2.5 = 3 – (0.2)r
$$ \Rightarrow $$ r = 2.5 $$\Omega $$
Power dissipated in internal resistance
= i2r = (0.2)2(2.5) = 0.1 W
$$ \Rightarrow $$ i2R = 0.5 W
iR = 2.5
$$ \Rightarrow $$ i = 0.2 A & R = 12.5 $$\Omega $$
Also, V = E – ir
$$ \Rightarrow $$ 2.5 = 3 – (0.2)r
$$ \Rightarrow $$ r = 2.5 $$\Omega $$
Power dissipated in internal resistance
= i2r = (0.2)2(2.5) = 0.1 W
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