Exercise EE1-02.05 – Current of a refrigerator (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-10-21)

Relates to lesson EE1-02 Electrical quantities and units, terms, symbols and models

Task

What current flows in the supply cable of a typical household refrigerator that draws the power $P=80\mathrm{W}$?

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Solution

Given

Given explicitly:

• Power $P=80\mathrm{W}$

Given implicitly from ‘typical household’:

• Mains voltage $U=230\mathrm{V}$ (standardized throughout Europe)
• Mains frequency $f=50\mathrm{Hz}$ (standardized throughout Europe, but irrelevant for the task)
• Maximum current $I_{\max }=16\mathrm{A}$ (because of the typical circuit breaker, but irrelevant for the task)

Required

Current $I$ in amperes ($\mathrm{A}$)

Solution approach

Approach: formula for calculating the power from voltage and current

$$P=U\cdot I$$

where:

• $P$ the power in watts ($\mathrm{W}$)
• $U$ the voltage in volts ($\mathrm{V}$)
• $I$ the current in amperes ($\mathrm{A}$)

Rearrange for $I$ and calculate the value:

$$I=\frac{P}{U}=\frac{80\mathrm{W}}{230\mathrm{V}}=0.347826087\mathrm{A}\approx \underline{0.35\mathrm{A}}$$

Answer

Through the supply cable of a refrigerator that draws $80\mathrm{W}$ flows a current of $I\approx 0.35\mathrm{A}$.

Precision of the result

Why is the result given as $0.35\mathrm{A}$ and not as $348\mathrm{mA}$ or $0.347826087\mathrm{A}$?
The precision of the given values determines the precision of the result. A chain is only as strong as its weakest link.

The voltage $U=230\mathrm{V}$ is given with three significant digits. The power value $P=80\mathrm{W}$ has two significant digits. If it were given as $80.0\mathrm{W}$, there would be three, but as it is, there are only two.

The result cannot be more precise than the least precise of the given values, which is why it is given with two significant digits. To do this, the last digit that remains is treated according to the rules of rounding, here: rounded up, because the digit that follows it is $7>5$. So the result is $0.35\mathrm{A}$.

This in turn could be expressed as $350\mathrm{mA}$, but this introduces an error, since the third significant digit, here the zero, suggests a precision that the given values cannot provide. An electrical engineer could live with this well in practice, because the value $350\mathrm{mA}$, with the SI prefix ‘milli’, contains a valuable hint at the order of magnitude $10^{-3}$, i.e. a rather small current. In (academic) studies, however, we try to be as precise as possible, which is why $\underline{0.35\mathrm{A}}$ is correct.