In the circuit shown the average power developed in the resistor `R_(1)` is

Options

A. 31.25 W
B. 62.50 W
C. 125 W
D. 250 W

Answer

B

Explanation

Rms current through resistor `R_(1)` is is
`l_(1(rms))=(50)/(sqrt(2)xx10sqrt(2))=(5)/(2)A`
`therefore` Average power developed in the resistor `R_(1)` is
`P_(1)=l_(1(rms))^(2)R_(1)=((5)/(2))^(2)10=(250)/(4)=62.50W`