Question : An alternating voltage V=30sin50t+40cos50t is applied to a resistor of resistance 10 Ω. The rms value of current through resistor is
(a) 5/√2 A
(b) 5√2 A
(c) 10/√2 A
(d) 10√2 A
Doubt by Gurdev
Solution :
In order to solve this question, we apply some identities which are given below
Solution :
V₀=√[A²+B²]
V₀=√[(30)²+(40)²]
V₀=√[900+1600]
V₀=√[2500]
V₀=50 V
V₀=√[A²+B²]
V₀=√[(30)²+(40)²]
V₀=√[900+1600]
V₀=√[2500]
V₀=50 V
R=10 Ω
V₀=I₀R
I₀=V₀/R
I₀=50/10
I₀=5 A
Irms=I₀/√2
Irms=5/√2 A
Hence, (a) 5/√2 A, would be the correct option.
V₀=I₀R
I₀=V₀/R
I₀=50/10
I₀=5 A
Irms=I₀/√2
Irms=5/√2 A
Hence, (a) 5/√2 A, would be the correct option.
