An alternating voltage given by V = 140 sin314t is connected across a pure resistor of 50 Ω, the rms current through the resistor is

Question

An alternating voltage given by V = 140 sin314t is connected across a pure resistor of 50 Ω, the rms current through the resistor is (A) 1.98 A (B) 5.63 A (C) 8.43 A (D) 2.39 A

Tardigrade Admin · Accepted Answer

Here R=50 Ω, V0=140 V ∴ Irms=(Vrms/R) =(0.707 V0/R) =(0.707 × 140/50) =1.98 A.

Q. An alternating voltage given by $V = 140 s in 314 t$ is connected across a pure resistor of $50 Ω$ , the rms current through the resistor is

$1.98 A$

$5.63 A$

$8.43 A$

$2.39 A$

Here $R = 50 Ω$ ,
$V_{0} = 140 V$
$∴ I_{r m s} = \frac{V _{r m s}}{R}$
$= \frac{0.707 V _{0}}{R}$
$= \frac{0.707 \times 140}{50}$
$= 1.98 A$ .

Q. An alternating voltage given by V=140sin314t is connected across a pure resistor of 50Ω, the rms current through the resistor is

1.98A

5.63A

8.43A

2.39A