A resistor `R` is connected in series with a coil. The system is subjected to an `AC` supply of peak voltage `V_(0)`. If the peak voltages dropped across the resistor `R` and the coil are `V_(1)` and `V_(2)` respectively
(a) The powerr dissipated in the coil is
`(V_(0)^(2) - V_(1)^(2) - V_(2)^(2))/(2R)`
The power dissipated in the circuit is
`(V_(0)^(2) + V_(1)^(2) - V_(2)^(2))/(2R)`

Options

A. Only (a) is correct
B. Only (b) is correct
C. (a),(b) are wrong
D. (a),(b) are correct

Answer

D

Explanation

Power dissipated in the coil is
`P = iV_(2) cos phi` ….(1)
and `V_(0)^(2) = V_(1)^(2) + V_(2)^(2) + 2V_(1) V_(2) cos phi` …..(2)
and `i = (V_(1))/(R )` ….(3)
Solving `P = (V_(0)^(2) - V_(1)^(2) - V_(2)^(2))/(2R)`
Total powe dissipated in the circuit is
`P' = iV_(0) cos alpha, i = (V)/(R )`
And `cos alpha = (V_(1) + V_(r))/(V_(0)) = (V_(1) + V_(2) cos phi)/(V_(0))`
Solving `P' (V_(0)^(2) + V_(1)^(2) - V_(2)^(2))/(2R)`