An alternating emf of 200 V, 50 Hz is applied to an L-R circuit, having a resistance R of 10 Ω and an inductance L of 0·05 H connected in series…

Given: V = 200 V, f = 50 Hz, R = 10 Ω, L = 0.05 H. X_L = 2π f L = 2π × 50 × 0.05 = 15.7 Ω (1) Impedance: Z = √(R² + X_L²) = √(10² + 15.7²) ≈ 18.6 Ω (2) Current: I = V/Z = (200)/(18.6) ≈ 10.7 A Z = 18.6 Ω, I = 10.7 A

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Answer the following.

An alternating emf of 200 V, 50 Hz is applied to an L-R circuit, having a resistance R of 10 Ω and…

(i)[3]

An alternating emf of $200\text{ V}$, $50\text{ Hz}$ is applied to an L-R circuit, having a resistance $R$ of $10\ \Omega$ and an inductance $L$ of $0\cdot05\text{ H}$ connected in series. Calculate: (1) Impedance. (2) Current flowing in the circuit.

(ii)[2]

Draw a labelled graph showing the variation of inductive reactance ($X_L$) verses frequency ($f$).

Draw: Graph showing the variation of inductive reactance ($X_L$) verses frequency ($f$)

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Answer

Answer (i)

Formulas used: Inductive reactance: XL = 2*pi*f*L with L = 0.05 H, f = 50 Hz, so XL = 15.7 ohm Impedance of a series LCR circuit (leave out the reactance it doesn't have): Z = sqrt(R**2 + (XL - XC)**2) with R = 10 ohm, XL = 15.7 ohm, so Z = 18.6 ohm Impedance of a series LCR circuit (leave out the reactance it doesn't have): Z = sqrt(R**2 + (XL - XC)**2) with V = 200 V, Z = 18.62 ohm, so I = 10.7 A

Given: $V = 200\ \text{V}$, $f = 50\ \text{Hz}$, $R = 10\ \Omega$, $L = 0.05\ \text{H}$. $X_L = 2\pi f L = 2\pi \times 50 \times 0.05 = 15.7\ \Omega$ (1) Impedance: $Z = \sqrt{R^2 + X_L^2} = \sqrt{10^2 + 15.7^2} \approx 18.6\ \Omega$ (2) Current: $I = \dfrac{V}{Z} = \dfrac{200}{18.6} \approx 10.7\ \text{A}$

Final answer: $Z = 18.6\ \Omega$, $I = 10.7\ \text{A}$

Answer (ii)

What it shows: $X_L = 2\pi f L$, so $X_L$ is directly proportional to $f$: a straight line through the origin with slope $2\pi L$.

$X_L = 2\pi f L$ is directly proportional to $f$, so the graph is a straight line through the origin with slope $2\pi L$ (drawn here for $L = 0.05$ H).