When an AC source of emf is connected across a circuit. The phase difference between the emf and the current in the circuit is observed to be , as shown in the diagram. If the circuit consists possibly only of or or in series, find the relationship

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. If the circuit consists possibly only of $R - C$ or $R - L$ or $L - C$ in series, find the relationship between the two elements. [2003]

Smart Achievers · Accepted Answer

(1)

$From e = E_{0} \sin (100 t),$ $∴$ $ω = 100$

From the given graph, current leads emf. Hence this is an $R - C$ circuit.

$\tan ϕ = \frac{X_{C}}{R}$

Here, $ϕ = 45 ° = \frac{π}{4}$ $∴$ $X_{C} = R$

$\frac{1}{ω C} = R \Rightarrow R C ω = 1$

or $R C = \frac{1}{100} s^{- 1}$ $∴$ $R = 1 k Ω, C = 10 μ F$

Solution

1 $R = 1 k Ω, C = 10 μ F$

2 $R = 1 k Ω, C = 1 μ F$

3 $R = 1 k Ω, L = 10 H$

4 $R = 1 k Ω, L = 1 H$

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Solution

1 R=1 kΩ, C=10 μF

2 R=1 kΩ, C=1 μF

3 R=1 kΩ, L=10 H

4 R=1 kΩ, L=1 H

Ans. (1) From e=E0sin(100t), ∴ ω=100 From the given graph, current leads emf. Hence this is an R-C circuit. tanϕ=XCR Here, ϕ=45°=π4 ∴ XC=R 1ωC=R ⇒ RCω=1 or RC=1100 s-1 ∴ R=1 kΩ, C=10 μF

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1 $R = 1 k Ω, C = 10 μ F$

2 $R = 1 k Ω, C = 1 μ F$

3 $R = 1 k Ω, L = 10 H$

4 $R = 1 k Ω, L = 1 H$