In the circuit shown in figure, the resistance R can be varied. By changing R power dissipated in it is made maximum. If the emf of the battery is 10 volt and internal resistance is 1 ohm then find the maximum power dissipated in R.
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a
10 W
b
15 W
c
20 W
d
25 W
answer is D.
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Detailed Solution
When power dissipated in R is maximum, R=r=1 Ω∴ i=εR+r=101+1A=5 A∴ Maximum power in R=(i)2R=(5)2×1 W=25 W