Question

An e.m.f. E = E₀ sin `omega`t is applied to a circuit containing 'L' and 'R' in series. If X_L = R, then the power dissipated in the circuit is ____________.

Options

• `"E"_0^2/(4"R")`

• `"E"_0/(4"R")`

• `"E"_0^2/(2"R")`

• `"E"_0/(2"R")`

Answer 1

An e.m.f. E = E₀ sin `omega`t is applied to a circuit containing 'L' and 'R' in series. If X_L = R, then the power dissipated in the circuit is `"E"_0^2/(4"R")`.

Explanation:

`"E" = "E"_0 "sin"  omega"t"`

`"P" = "E"_0/sqrt2 xx "I"_0/sqrt2  "cos" theta`

`= "E"_0/sqrt2 xx "E"_0/sqrt(2"Z") xx "R"/"Z"`

`= ("E"_0^2"R")/(2"Z"^2)       "Z"^2 = "R"^2 + omega"L"^2 = 2"R"^2`

`= ("E"_0^2 "R")/(4"R"^2)`

`= "E"_0^2/(4"R")`