When an AC source of emf e-E0sin100t is connected across a circuit- the phase difference between the emf e and the current i in the circuit is observed to be -4- as shown in the diagram- lf the circuit consists possibly only of R-C or R-L or L-C in series- find the relationship between the two elements-

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RLC Circuit

When an AC so...

Question

When an AC source of emf $e = E_{0} sin (100 t)$ is connected across a circuit, the phase difference between the emf $e$ and the current $i$ in the circuit is observed to be $π / 4$ , as shown in the diagram. lf the circuit consists possibly only of $R - C$ or $R - L$ or $L - C$ in series, find the relationship between the two elements.

R = 1 k Ω, C = 10 μ F

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R = 1 k Ω, C = 1 μ F

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R = 1 k Ω, L = 10 H

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R = 1 k Ω, L = 1 H

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Solution

The correct option is A $R = 1 k Ω, C = 10 μ F$
Here, the current is ahead of potential. Thus the circuit is R-C circuit.
Since the phase difference is $ϕ = π / 4$ ,

$t a n ϕ = 1 = \frac{1}{R ω C}$

$⟹ R = \frac{1}{ω C}$

Thus only option A satisfies the condition.

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The correct option is A R=1 kΩ,C=10μFHere, the current is ahead of potential. Thus the circuit is R-C circuit.Since the phase difference is ϕ=π/4, tanϕ=1=1RωC⟹R=1ωCThus only option A satisfies the condition.

The correct option is A $R = 1 k Ω, C = 10 μ F$
Here, the current is ahead of potential. Thus the circuit is R-C circuit.
Since the phase difference is $ϕ = π / 4$ ,

$t a n ϕ = 1 = \frac{1}{R ω C}$

$⟹ R = \frac{1}{ω C}$

Thus only option A satisfies the condition.