SSC JE Electrical Engineering 2015 Paper

Question : 109 of 199

Marks: +1, -0

The internal resistance of a voltage source is 10 Ω and has 10 volts at its terminals. Find the maximum power that can be transferred to the load.

25 W
5 W
0.25 W
2.5 W

Validate

Solution:

For maximum power to be transferred from source to a load Resistance, the value of load Resistance should be equal to the internal Resistance of the source

R_L = R_{\th }= 10Ω

P_L={{V^2}_{\th}}/{4R_L}=(10)^2/{4×10}=2.5

watt.

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