A cell of emf 8.0 V having a finite resistance of 4Ω is connected to a load resistance of xΩ for maximum power transfer. The value of x should be
A
0 ohm
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B
4 ohm
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C
3 ohm
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D
2 ohm
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Solution
The correct option is B 4 ohm P=i2.x =(∈4+x)2.x for Pmax,dPdx=0 dpdx=ddx{∈2.x(4+x)2}=∈2ddx{x(4+x)2} dpdx=0=(4+x)2.ddx(x)−xddx(4+x)2(4+x)4 0=(4+x)2−2x(4+x) 0=(4+x)(4−x)