Gives
R1=4ΩR2=6ΩR3=10ΩR4=15Ω
R1 and R2 net resistance R'1
R1′=R1+R2R1R2=4+64×6=1024=512Ω
R3 and R4 net resistance R'2
R2′=R3+R4R3R4=10+1510×15=25150=6Ω
Rnet=R1′+R2′R1′R2′=512+6512×6=1.71Ω
P=IV
Potential is same all resistance
Then P high ,I High
4 ohm resistance Power maximum
Pmax=IV
Pmax=I2R=4I2watt
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