When an A C source of emf e=E0 sin (100t) is connected across a circuit, the phase difference between the emf e and the current i in the circuit is observed to be (π/4) ahead, as shown in the diagram. If the circuit consists possibly only of R-C or R-L or L-C in series, find the relationship between the two elements.

Question

When an A C source of emf e=E0 sin (100t) is connected across a circuit, the phase difference between the emf e and the current i in the circuit is observed to be (π/4) ahead, as shown in the diagram. If the circuit consists possibly only of R-C or R-L or L-C in series, find the relationship between the two elements. (A) R= 1 k Ω, C = 10 μ F (B) R= 1 k Ω, C = 1 μ F (C) R= 1 k Ω, L = 10 H (D) R= 1 k Ω, L = 1 H

Tardigrade Admin · Accepted Answer

Here, the current is ahead of potential. Thus the circuit is R-C circuit. Since the phase difference is φ=π / 4, tan φ=1=(1/R ω C) Longrightarrow R =(1/ω C )

$R = 1 k Ω, C = 10 μ F$

$R = 1 k Ω, C = 1 μ F$

$R = 1 k Ω, L = 10 H$

$R = 1 k Ω, L = 1 H$

Here, the current is ahead of potential. Thus the circuit is R-C circuit. Since the phase difference is $ϕ = π /4$ ,
$tan ϕ = 1 = \frac{1}{R ω C}$
$⟹ R = \frac{1}{ω C}$

R=1kΩ,C=10μF

R=1kΩ,C=1μF

R=1kΩ,L=10H

R=1kΩ,L=1H

Here, the current is ahead of potential. Thus the circuit is R-C circuit. Since the phase difference is ϕ=π/4, tanϕ=1=RωC1​ ⟹R=ωC1​

$R = 1 k Ω, C = 10 μ F$

$R = 1 k Ω, C = 1 μ F$

$R = 1 k Ω, L = 10 H$

$R = 1 k Ω, L = 1 H$

Here, the current is ahead of potential. Thus the circuit is R-C circuit. Since the phase difference is $ϕ = π /4$ ,
$tan ϕ = 1 = \frac{1}{R ω C}$
$⟹ R = \frac{1}{ω C}$