Exercise EE1-03.R2 (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-12-09)

Relates to EE1-03 Sources

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Task

An audio amplifier delivers a maximum output power of $P_{\mathrm{out}}=100\mathrm{W}$ to a loudspeaker. The amplifier has an efficiency of $\eta =70\%$.

a) What electrical input power $P_{\mathrm{in}}$ must be supplied to the amplifier?

b) What power loss $P_{\mathrm{loss}}$ occurs in the amplifier?

c) The amplifier is supplied internally by a power supply with a direct voltage of $U=36\mathrm{V}$. What current $I$ must the power supply deliver when the amplifier delivers the maximum output power?

d) What is the maximum permissible internal resistance $R_i$ of the power supply if the supply voltage is not to dip by more than $5\%$ at full load?

Solution

a) $P_{\mathrm{in}}\approx 143\mathrm{W}$
b) $P_{\mathrm{loss}}\approx 43\mathrm{W}$
c) $I\approx 3.97\mathrm{A}$
d) $R_i\le 0.45\Omega$

Full solution

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Solution

Given

Given explicitly:

• Output power: $P_{\mathrm{out}}=100\mathrm{W}$
• Efficiency: $\eta =70\%=0.7$
• Supply voltage (open circuit): $U_0=36\mathrm{V}$
• Maximum voltage dip: $5\%$

Known:

• Efficiency: $\eta =\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}$
• Power loss: $P_{\mathrm{loss}}=P_{\mathrm{in}}-P_{\mathrm{out}}$
• Power: $P=U\cdot I$
• Linear voltage source: $U_{\mathrm{t}}=U_0-R_i\cdot I$

where:

• $\eta$ is the efficiency (dimensionless or in $\%$)
• $P_{\mathrm{out}}$ is the output power (useful output power) in watts ($\mathrm{W}$)
• $P_{\mathrm{in}}$ is the input power (power supplied) in watts ($\mathrm{W}$)
• $P_{\mathrm{loss}}$ is the power loss in watts ($\mathrm{W}$)
• $U_0$ is the open-circuit voltage in volts ($\mathrm{V}$)
• $U_{\mathrm{t}}$ is the terminal voltage in volts ($\mathrm{V}$)
• $R_i$ is the internal resistance in ohms ($\Omega$)

Required

a) Input power $P_{\mathrm{in}}$ in $\mathrm{W}$
b) Power loss $P_{\mathrm{loss}}$ in $\mathrm{W}$
c) Current draw $I$ in $\mathrm{A}$
d) Maximum internal resistance $R_i$ in $\Omega$

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a) Input power

From the definition of efficiency:

$$\eta =\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}\Rightarrow P_{\mathrm{in}}=\frac{P_{\mathrm{out}}}{\eta }$$ $$P_{\mathrm{in}}=\frac{100\mathrm{W}}{0.7}=142.86\mathrm{W}\approx \underline{143\mathrm{W}}$$

b) Power loss

The difference between the power supplied and the power delivered is the power loss:

$$P_{\mathrm{loss}}=P_{\mathrm{in}}-P_{\mathrm{out}}=142.86\mathrm{W}-100\mathrm{W}=42.86\mathrm{W}\approx \underline{43\mathrm{W}}$$

Check: An efficiency of $70\%$ means that $30\%$ of the input power is lost:

$$P_{\mathrm{loss}}=(1-\eta )\cdot P_{\mathrm{in}}=0.3\cdot 142.86\mathrm{W}=42.86\mathrm{W}$$

✔

Heat dissipation

The power loss of approx. $43\mathrm{W}$ is converted into heat in the amplifier. This heat must be dissipated via heat sinks or fans to prevent overheating.

c) Current draw

From the input power of the amplifier at full load determined in a) and the supply voltage, you can determine the required current that flows from the power supply to the amplifier:

$$P_{\mathrm{in}}=U\cdot I\Rightarrow I=\frac{P_{\mathrm{in}}}{U}$$ $$I=\frac{142.86\mathrm{W}}{36\mathrm{V}}=3.97\mathrm{A}\approx \underline{4.0\mathrm{A}}$$

d) Maximum internal resistance

With a voltage dip of at most $5\%$, the terminal voltage must be at least:

$$U_{\mathrm{t},\min }=U_0\cdot (1-0.05)=36\mathrm{V}\cdot 0.95=34.2\mathrm{V}$$

The maximum voltage drop across the internal resistance:

$$\Delta U=U_0-U_{\mathrm{t},\min }=36\mathrm{V}-34.2\mathrm{V}=1.8\mathrm{V}$$

Using Ohm’s law, the maximum internal resistance follows:

$$R_i=\frac{\Delta U}{I}=\frac{1.8\mathrm{V}}{3.97\mathrm{A}}=0.453\Omega \approx \underline{0.45\Omega }$$

Check via the terminal voltage:

$$U_{\mathrm{t}}=U_0-R_i\cdot I=36\mathrm{V}-0.453\Omega \cdot 3.97\mathrm{A}=36\mathrm{V}-1.8\mathrm{V}=34.2\mathrm{V}$$

✔

Practical significance

An internal resistance of $R_i\le 0.45\Omega$ places high demands on the power supply. Laboratory power supplies typically have internal resistances in the range of $10$ to $100\mathrm{m\Omega }$ and therefore meet this requirement easily. Simple plug-in power supplies, however, can have considerably higher internal resistances.