Exercise EE1-02.04 – Case fan schematic, symbols, calculation (Solution)

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

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

Task

A case fan is designed for connection to $12\mathrm{V}$ DC. A series resistor reduces the operating voltage of the fan to $9.0\mathrm{V}$ so that the fan runs more slowly and more quietly. In low-noise operation, a current of $40\mathrm{mA}$ is measured.

a) Sketch the electric circuit and enter all electrical quantities.

b) Calculate the resistance of the fan at this operating point and enter the value in the schematic as well. Here, you only calculate the resistance of the fan in operation, without the series resistor.

c) Using the calculated fan resistance, calculate the current for operation without the series resistor (at $12\mathrm{V}$).

◀️ Back to the task

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Solution

Given

Given explicitly:

• Supply voltage $U_{\mathrm{tot}}=12.0\mathrm{V}$

• Voltage across the fan (reduced) $U_{\mathrm{fan}}=9.0\mathrm{V}$

• Current in low-noise operation $I=40\mathrm{mA}$

Given implicitly:

• Voltage across the series resistor $U_{\mathrm{ser}}=U_{\mathrm{tot}}-U_{\mathrm{fan}}=12\mathrm{V}-9.0\mathrm{V}=3.0\mathrm{V}$

Required

a) Circuit sketch with the quantities entered
b) Resistance of the fan $R_{\mathrm{fan}}$ in ohms ($\Omega$)
c) Current $I_{12\mathrm{V}}$ with direct connection to $12\mathrm{V}$ in amperes ($\mathrm{A}$)

a) Circuit sketch

The electric circuit consists of a voltage source ($12\mathrm{V}$) to which the series connection of the series resistor $R_{\mathrm{ser}}$ and the fan $R_{\mathrm{fan}}$ is connected.

Quantities entered:

• $U_{\mathrm{tot}}=12\mathrm{V}$ (supply voltage)
• $U_{\mathrm{fan}}=9.0\mathrm{V}$ (voltage across the fan)
• $I=40\mathrm{mA}$ (current through the series connection)
• $R_{\mathrm{ser}}$ (series resistor)
• $R_{\mathrm{fan}}$ (fan resistance)

b) Resistance of the fan

According to Ohm’s law:

$$R=\frac{U}{I}$$

Resistance of the fan:

$$R_{\mathrm{fan}}=\frac{U_{\mathrm{fan}}}{I}=\frac{9.0\mathrm{V}}{40\mathrm{mA}}=\frac{9.0\mathrm{V}}{0.040\mathrm{A}}=225\Omega \approx \underline{2.3\cdot 10^2\Omega }$$

The resistance of the fan is $225\Omega$ or, rounded to the correct precision, $2.3\cdot 10^2\Omega$.

Calculating the series resistor

From the voltage across the series resistor, you can also calculate its value:

$$R_{\mathrm{ser}}=\frac{U_{\mathrm{ser}}}{I}=\frac{3.0\mathrm{V}}{0.040\mathrm{A}}=75\Omega$$

c) Current with direct connection to 12 V

Without the series resistor, the full voltage is applied to the fan:

$$I_{12\mathrm{V}}=\frac{U_{\mathrm{tot}}}{R_{\mathrm{fan}}}=\frac{12\mathrm{V}}{225\Omega }=0.0533\mathrm{A}=53.3\mathrm{mA}\approx \underline{53\mathrm{mA}}$$

With direct connection to $12\mathrm{V}$, a current of $53\mathrm{mA}$ flows through the fan.

Rotational frequency and loudness

The current increases from $40\mathrm{mA}$ to $53\mathrm{mA}$ (by $33\%$). This leads to a higher rotational frequency and thus to louder operation. The series resistor throttles the fan so that it runs more quietly – however, the excess power of about $0.12\mathrm{W}$ is converted into heat in the series resistor.