Exercise EE1-04.R2 (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-12-07)

Relates to EE1-04 Resistance

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Task

An electric heating element made of nickel-chromium wire (NiCr) is used in an industrial furnace. At room temperature ($T_1=20°\mathrm{C}$), the resistance of the heating element is $R_{20}=48\Omega$.

In operation, the heating wire heats up to a temperature of $T_2=800°\mathrm{C}$.

The temperature coefficient of NiCr is ${\alpha }_{20}=0.4\cdot 10^{-3}{\mathrm{K}}^{-1}$.

a) Calculate the resistance of the heating element at operating temperature.

b) The heating element is operated at a voltage of $U=230\mathrm{V}$. Calculate the current and the power in the cold state (at $20°\mathrm{C}$).

c) Calculate the current and the power in the hot state (at $800°\mathrm{C}$).

d) By what percentage does the power consumption change between the cold and the hot state?

Solution

a) $R_{800}=62.98\Omega \approx 63\Omega$
b) $I_{\mathrm{cold}}=4.79\mathrm{A}$, $P_{\mathrm{cold}}=1102\mathrm{W}$
c) $I_{\text{hot}}=3.65\mathrm{A}$, $P_{\text{hot}}=840\mathrm{W}$
d) Decrease in power of approx. $24\%$

Full solution

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Solution

Given

Given explicitly:

• Resistance at room temperature: $R_{20}=48\Omega$
• Room temperature: $T_1=20°\mathrm{C}$
• Operating temperature: $T_2=800°\mathrm{C}$
• Temperature coefficient of NiCr: ${\alpha }_{20}=0.4\cdot 10^{-3}{\mathrm{K}}^{-1}$
• Operating voltage: $U=230\mathrm{V}$

Known:

• Temperature dependence of the resistance: $R(T)=R_{20}\cdot (1+{\alpha }_{20}\cdot \Delta T)$
• Temperature difference: $\Delta T=T_2-T_1$
• Ohm’s law: $I=\frac{U}{R}$
• Power: $P=U\cdot I=\frac{U^2}{R}$

where:

• $R(T)$ the resistance at temperature $T$ in ohms ($\Omega$)
• $R_{20}$ the resistance at $20°\mathrm{C}$ (reference temperature) in ohms ($\Omega$)
• ${\alpha }_{20}$ the temperature coefficient at $20°\mathrm{C}$ in ${\mathrm{K}}^{-1}$
• $\Delta T$ the temperature difference from the reference temperature in kelvins ($\mathrm{K}$)
• $I$ the current in amperes ($\mathrm{A}$)
• $P$ the power in watts ($\mathrm{W}$)

Required

a) Resistance $R_{800}$ at $800°\mathrm{C}$
b) Current $I_{\mathrm{cold}}$ and power $P_{\mathrm{cold}}$ at $20°\mathrm{C}$
c) Current $I_{\text{hot}}$ and power $P_{\text{hot}}$ at $800°\mathrm{C}$
d) Relative change in power in per cent

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a) Resistance at operating temperature

The temperature difference is:

$$\Delta T=T_2-T_1=800°\mathrm{C}-20°\mathrm{C}=780\mathrm{K}$$

The resistance at $800°\mathrm{C}$:

$$R_{800}=R_{20}\cdot (1+{\alpha }_{20}\cdot \Delta T)$$ $$R_{800}=48\Omega \cdot \left(1+0.4\cdot 10^{-3}{\mathrm{K}}^{-1}\cdot 780\mathrm{K}\right)$$ $$R_{800}=48\Omega \cdot (1+0.312)=48\Omega \cdot 1.312=\underline{62.98\Omega }$$

Note on the choice of material

NiCr (nickel-chromium, also known as Nichrome) has a particularly low temperature coefficient (${\alpha }_{20}\approx 0.4\cdot 10^{-3}{\mathrm{K}}^{-1}$) compared with pure metals such as copper (${\alpha }_{20}\approx 3.93\cdot 10^{-3}{\mathrm{K}}^{-1}$). This makes NiCr ideal for heating elements, since the change in resistance remains moderate at high temperatures.

b) Current and power in the cold state

Current at $20°\mathrm{C}$:

$$I_{\mathrm{cold}}=\frac{U}{R_{20}}=\frac{230\mathrm{V}}{48\Omega }=\underline{4.79\mathrm{A}}$$

Power at $20°\mathrm{C}$:

$$P_{\mathrm{cold}}=U\cdot I_{\mathrm{cold}}=230\mathrm{V}\cdot 4.79\mathrm{A}=\underline{1102\mathrm{W}}$$

Check using $P=U^2/R$:

$$P_{\mathrm{cold}}=\frac{(230\mathrm{V}{)}^2}{48\Omega }=\frac{52900{\mathrm{V}}^2}{48\Omega }=1102\mathrm{W}$$

✔

c) Current and power in the hot state

Current at $800°\mathrm{C}$:

$$I_{\text{hot}}=\frac{U}{R_{800}}=\frac{230\mathrm{V}}{62.98\Omega }=\underline{3.65\mathrm{A}}$$

Power at $800°\mathrm{C}$:

$$P_{\text{hot}}=U\cdot I_{\text{hot}}=230\mathrm{V}\cdot 3.65\mathrm{A}=\underline{840\mathrm{W}}$$

Check using $P=U^2/R$:

$$P_{\text{hot}}=\frac{(230\mathrm{V}{)}^2}{62.98\Omega }=\frac{52900{\mathrm{V}}^2}{62.98\Omega }=840\mathrm{W}$$

✔

d) Relative change in power

The relative change in power is:

$$\frac{\Delta P}{P_{\mathrm{cold}}}=\frac{P_{\mathrm{cold}}-P_{\text{hot}}}{P_{\mathrm{cold}}}=\frac{1102\mathrm{W}-840\mathrm{W}}{1102\mathrm{W}}$$ $$\frac{\Delta P}{P_{\mathrm{cold}}}=\frac{262\mathrm{W}}{1102\mathrm{W}}=0.238=\underline{23.8\%}$$

The power consumption decreases by about 24 % when the heating element reaches its operating temperature.

Practical consequence

This behaviour is typical of resistive heating elements: at switch-on, a higher inrush current flows, which decreases as the element heats up. The control of industrial furnaces must take this behaviour into account. With incandescent bulbs, the effect is even more drastic (tungsten wire has ${\alpha }_{20}\approx 4.5\cdot 10^{-3}{\mathrm{K}}^{-1}$), which is why the inrush current there can be about 10 to 15 times the operating current.