Exercise EE1-04.02 – Power of a resistor (Solution)

Prof. Dr. Thorsten Jungmann (last updated 2025-10-27)

Relates to EE1-04 Resistance

Task

a) What maximum current $I_{max}$ may flow through a resistor $R=485\Omega$ that is rated for a maximum permissible power $P_{max}=10\mathrm{W}$? What voltage is then dropped across the resistor?

b) For what maximum power $P_{max}$ must a resistor be rated at least if, at an applied voltage $U=42\mathrm{V}$, a current $I=0.23\mathrm{A}$ flows through it?

c) What maximum voltage $U_{max}$ may be applied to a resistor through which a current $I=130\mathrm{mA}$ flows if it is rated for a maximum power $P_{max}=250\mathrm{mW}$?

Solution

a) $0.14\mathrm{A}$; $70\mathrm{V}$
b) $10\mathrm{W}$
c) $1.92\mathrm{V}$

▶️ Full worked solution

◀️ Back to the task

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Solution

Formulas: $P=U\cdot I=\frac{U^2}{R}=I^{2}R$

a) Given: $R=485\Omega ,P_{\max }=10\mathrm{W}$ – Required: $I_{\max },U$

$$I_{\max }=\sqrt{\frac{P_{\max }}{R}}=\sqrt{\frac{10\mathrm{W}}{485\Omega }}=0.1436\mathrm{A}\approx \underline{0.14\mathrm{A}}$$ $$U=I_{\max }R=\sqrt{P_{\max }R}=\sqrt{10\mathrm{W}\cdot 485\Omega }=69.64\mathrm{V}\approx \underline{70\mathrm{V}}$$

b) Given: $U=42\mathrm{V},I=0.23\mathrm{A}$ – Required: $P_{\max }$ (minimum rating)

$$P=U\cdot I=42\mathrm{V}\cdot 0.23\mathrm{A}=9.66\mathrm{W}\Rightarrow \text{minimum power rating: }\underline{10\mathrm{W}}$$

c) Given: $I=130\mathrm{mA}=0.130\mathrm{A},P_{\max }=250\mathrm{mW}=0.250\mathrm{W}$ – Required: $U_{\max }$

$$U_{\max }=\frac{P_{\max }}{I}=\frac{0.250\mathrm{W}}{0.130\mathrm{A}}=1.923\mathrm{V}\approx \underline{1.92\mathrm{V}}$$