---
title: "Induction furnace at 10 V, half-full hearth"
subject: "Utilization of Electrical Energy"
chapter: 5
key: "semester_7/utilization_of_electrical_energy:induction-furnace-10v-halved-resistance"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/induction-furnace-10v-halved-resistance
site: BCT Engineering
---

# Induction furnace at 10 V, half-full hearth

Utilization of Electrical Energy · Chapter 5: Electric Heating

- Asked in: 2073 Shrawan
- Video: [Induction Heating (Basics, Working, Circuit, Applications, Block Diagram & Applications) Explained](https://www.youtube.com/watch?v=8KPYYmNMR5o) — Engineering Funda
- 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-10v-halved-resistance
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-5

## Question

A low frequency induction furnace operating at 10V in the secondary circuit takes 500kW at 0.5 p.f when the hearth is full. If the secondary voltage be maintained at 10V, estimate the power absorbed and the p.f when the hearth is half full. Assume the resistance of the secondary circuit to be thereby halved and the reactance to remain the same.

## Short answer

Full hearth gives R = 0.5×10⁻⁴ ohm and X = 0.866×10⁻⁴ ohm. Half full gives Z' = 0.9014×10⁻⁴ ohm and I' = 110,940 A, so power is about 307.7 kW and pf is 0.277 lagging.

## Answer

In a low-frequency core-type induction furnace the charge forms a single-turn secondary, so its current depends on the secondary resistance $R$ and reactance $X$.

**Given:** $V_2 = 10$ V, $P = 500$ kW at pf 0.5 (hearth full); half-full: $R' = R/2$, $X$ unchanged.

#### Hearth full
$$
\begin{aligned}
S &= \frac{500}{0.5} = 1000\ \text{kVA}, \quad I_2 = \frac{1000\times10^3}{10} = 100\,000\ \text{A} \\
Z &= \frac{10}{100\,000} = 1.0 \times 10^{-4}\ \Omega \\
R &= Z\cos\phi = 0.5\times10^{-4}\ \Omega, \quad X = Z\sin\phi = 0.866 \times 10^{-4}\ \Omega
\end{aligned}
$$

#### Hearth half full
$$
\begin{aligned}
R' &= 0.25 \times 10^{-4}\ \Omega \\
Z' &= \sqrt{0.25^2 + 0.866^2} \times 10^{-4} = 0.9014 \times 10^{-4}\ \Omega \\
I_2' &= \frac{10}{0.9014 \times 10^{-4}} = 110\,940\ \text{A} \\
P' &= I_2'^2 R' = 110\,940^2 \times 0.25 \times 10^{-4} = 307.7\ \text{kW} \\
\cos\phi' &= \frac{R'}{Z'} = \frac{0.25}{0.9014} = 0.277\ \text{lagging}
\end{aligned}
$$

**Answer:** Power absorbed $\approx 307.7$ kW at power factor $\approx 0.277$ lagging. Halving the resistance raises the current but lowers both the power and the power factor, because the circuit becomes mainly reactive.

## Check yourself

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

1. What are Z and R for the full hearth (10 V, 500 kW, 0.5 pf)?
   A. Z = 2x10^-4 Ω, R = 1x10^-4 Ω
   B. Z = 0.5x10^-4 Ω, R = 1x10^-4 Ω
   C. Z = 1x10^-4 Ω, R = 0.866x10^-4 Ω
   D. Z = 1.0x10^-4 Ω, R = 0.5x10^-4 Ω
   Answer: D. I2 = 1000 kVA/10 V = 100,000 A, Z = 10^-4 Ω, R = Z cos(phi) = 0.5x10^-4 Ω. (Key point: Full-hearth impedance)

2. When R is halved, what happens to current and power?
   A. Both rise
   B. Both fall
   C. Current rises to about 110,940 A but power falls to about 307.7 kW
   D. Power doubles
   Answer: C. Z' falls to 0.9014x10^-4 Ω so I rises, but P' = I'^2 R' with half R gives 307.7 kW. (Key point: Effect of halving R)

3. Why does the power factor fall to about 0.277?
   A. X is halved
   B. Circuit becomes mainly reactive as R decreases while X is unchanged
   C. Voltage falls
   D. The core is saturated
   Answer: B. cos(phi') = R'/Z' = 0.25/0.9014 = 0.277 lagging. (Key point: Power factor fall)

## Close variants set in other exams

- [Low frequency induction furnace with half-full hearth](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ainduction-furnace-half-full-hearth) — Utilization of Electrical Energy, chapter 5, asked 2×