---
title: "Low frequency induction furnace with half-full hearth"
subject: "Utilization of Electrical Energy"
chapter: 5
key: "semester_7/utilization_of_electrical_energy:induction-furnace-half-full-hearth"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/induction-furnace-half-full-hearth
site: BCT Engineering
---

# Low frequency induction furnace with half-full hearth

Utilization of Electrical Energy · Chapter 5: Electric Heating

- Asked in: 2082 Baisakh, 8 marks
- Video: [Induction Heating |Heating & Welding| |Utilization of Electrical Energy|](https://www.youtube.com/watch?v=j3Yb6FItwwY) — Anish Vora
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/induction-furnace-half-full-hearth
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-5

## Question

A low frequency induction furnace operating at 20 volts in secondary circuit takes 500 kW at 0.5 pf when the hearth is full. If the secondary voltage be maintained to 20 volts, estimate the power absorbed and the power factor when the hearth is half-full. Assume the resistance of the secondary circuit to be thereby doubled and the reactance to remain the same.

## Short answer

For low-frequency induction furnace at 20V secondary: full hearth 500kW at 0.5pf gives Z=4e-4Ω, R=2e-4Ω, X=3.464e-4Ω. Half-full: R'=4e-4Ω, X unchanged. New Z'=5.2915e-4Ω, I'=37,796A. Power absorbed = I'²R' ≈ 571.4 kW, power factor = R'/Z' = 0.756 lagging.

## Answer

In a low-frequency (core-type) induction furnace the charge forms the short-circuited secondary, so the secondary current is set by the secondary resistance $R$ and reactance $X$.

**Given:** $V_2 = 20$ V, $P = 500$ kW at pf 0.5 (hearth full). Half-full: $R' = 2R$, $X$ unchanged.

#### Secondary impedance when hearth is full
$$
\begin{aligned}
S &= \frac{P}{\cos\phi} = \frac{500}{0.5} = 1000\ \text{kVA} \\
I_2 &= \frac{S}{V_2} = \frac{1000 \times 10^3}{20} = 50\,000\ \text{A} \\
Z &= \frac{V_2}{I_2} = \frac{20}{50\,000} = 4 \times 10^{-4}\ \Omega \\
R &= Z\cos\phi = 2 \times 10^{-4}\ \Omega \\
X &= Z\sin\phi = 4\times10^{-4} \times 0.866 = 3.464 \times 10^{-4}\ \Omega
\end{aligned}
$$

#### Hearth half full
$$
\begin{aligned}
R' &= 2R = 4 \times 10^{-4}\ \Omega, \quad X = 3.464 \times 10^{-4}\ \Omega \\
Z' &= \sqrt{(4)^2 + (3.464)^2} \times 10^{-4} = 5.2915 \times 10^{-4}\ \Omega \\
I_2' &= \frac{20}{5.2915 \times 10^{-4}} = 37\,796\ \text{A} \\
P' &= I_2'^2 R' = 37\,796^2 \times 4 \times 10^{-4} = 571.4\ \text{kW} \\
\cos\phi' &= \frac{R'}{Z'} = \frac{4}{5.2915} = 0.756\ \text{lagging}
\end{aligned}
$$

**Answer:** With the hearth half full, power absorbed $\approx 571.4$ kW at a power factor of $0.756$ lagging.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For the full hearth (500 kW, 0.5 pf, 20 V), what are Z and R?
   A. Z = 2x10^-4 Ω, R = 4x10^-4 Ω
   B. Z = 4x10^-4 Ω, R = 3.464x10^-4 Ω
   C. Z = 4x10^-4 Ω, R = 2x10^-4 Ω
   D. Z = 8x10^-4 Ω, R = 4x10^-4 Ω
   Answer: C. S = 1000 kVA, I = 50,000 A, Z = 20/50,000 = 4x10^-4 Ω, and R = Z cos(phi) = 2x10^-4 Ω. (Key point: Finding R and X when full)

2. When the hearth is half full, which quantities change?
   A. Both R and X double
   B. R doubles, X unchanged
   C. R unchanged, X doubles
   D. Voltage doubles
   Answer: B. The problem states R doubles to 4x10^-4 Ω, X stays 3.464x10^-4 Ω, and V2 stays 20 V. (Key point: Half-full condition)

3. How is the power absorbed with half-full hearth found?
   A. P = V2 x I2 only
   B. P = I2'^2 R' with I2' = V2/Z', giving about 571 kW
   C. P = 500/2 = 250 kW
   D. P = I2^2 R using the full-hearth current
   Answer: B. Z' = 5.29x10^-4 Ω gives I2' = 37,796 A, and P' = I2'^2 x 4x10^-4 = 571.4 kW. (Key point: Power absorbed)

4. What is the new power factor?
   A. 0.5
   B. 0.866
   C. 0.756 lagging
   D. 1.0
   Answer: C. cos(phi') = R'/Z' = 4/5.2915 = 0.756 lagging. (Key point: New power factor)

## Close variants set in other exams

- [Induction furnace at 10 V, half-full hearth](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ainduction-furnace-10v-halved-resistance) — Utilization of Electrical Energy, chapter 5, asked 2×

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