SSC JE Electrical Engineering 2010 Paper

Question : 32 of 50

Marks: +1, -0

R_g

in the circuit shown in figure 1 is variable between 20 Ω and 80 Ω then maximum power transferred to the load

R_L

will be

Solution:

Here in the given circuit

R_g

= 20Ω to 80Ω
∵

R_g

should be minimum to transfer maximum power to the load (

R_L

), so we take

R_g=20Ω

∴I=V/{R_g+R_L}=40/{20+60}

=0.5A

∴P_\max =I^2R_L

=0.5^2×60=60/4=15W

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