When a voltage V=V0 cos ω t is applied across a resistor of resistance R, the average power dissipated per cycle in the resistor is given by

Question

When a voltage V=V0 cos ω t is applied across a resistor of resistance R, the average power dissipated per cycle in the resistor is given by (A) (V0/√2 R) (B) (V0/√2 ω R) (C) (V02/2 R) (D) (V02/2 ω R)

Tardigrade Admin · Accepted Answer

P=E textrms I textrms cos φ cos φ=(R/Z) cos φ=1 for resistor V textrms =(V0/√2) P=(V0/√2), (V0/R √2)(1) (∵ I textrms =(V textrms /R)=(V0/R √2)) P=(V02/2 R)

Q. When a voltage $V = V_{0} cos ω t$ is applied across a resistor of resistance $R$ , the average power dissipated per cycle in the resistor is given by

$\frac{V _{0}}{2 R}$

$\frac{V _{0}}{2 ω R}$

$\frac{V _{0}^{2}}{2 R}$

$\frac{V _{0}^{2}}{2 ω R}$

$P = E_{rms} I_{rms} cos ϕ$
$cos ϕ = \frac{R}{Z}$
$cos ϕ = 1$ for resistor
$V_{rms} = \frac{V _{0}}{2}$
$P = \frac{V _{0}}{2}, \frac{V _{0}}{R 2} (1)$
$(∵ I_{rms} = \frac{V _{rms}}{R} = \frac{V _{0}}{R 2})$
$P = \frac{V _{0}^{2}}{2 R}$

Q. When a voltage V=V0​cosωt is applied across a resistor of resistance R, the average power dissipated per cycle in the resistor is given by

2​RV0​​

2​ωRV0​​

2RV02​​

2ωRV02​​