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Difficulty: HardElectrical Power

In a direct current (DC) avionics test circuit, a 48 V48\text{ V} power supply is connected in series with a 30 Ω30\text{ }\Omega current-limiting resistor and a 50 Ω50\text{ }\Omega load resistor. What is the electrical power dissipated by the 50 Ω50\text{ }\Omega load resistor, in watts?

Answer: 18 W

Answer

The power dissipated by the 50 Ω50\text{ }\Omega load resistor is 18 W18\text{ W}.
In a series circuit, the total resistance is the sum of all individual resistances (30 Ω+50 Ω=80 Ω30\text{ }\Omega + 50\text{ }\Omega = 80\text{ }\Omega). With a 48 V48\text{ V} supply, the circuit current is I=48 V80 Ω=0.6 AI = \frac{48\text{ V}}{80\text{ }\Omega} = 0.6\text{ A}. The electrical power dissipated by the 50 Ω50\text{ }\Omega resistor is determined by P=I2R=(0.6 A)2×50 Ω=18 WP = I^2 R = (0.6\text{ A})^2 \times 50\text{ }\Omega = 18\text{ W}.

Step-by-Step Solution

1
Find the equivalent total resistance of the series circuit
Rtotal=30 Ω+50 Ω=80 ΩR_{\text{total}} = 30\text{ }\Omega + 50\text{ }\Omega = 80\text{ }\Omega
Resistances in a series circuit are directly additive to determine total opposition to current flow.
2
Calculate the total circuit current using Ohm's Law
I=VtotalRtotal=48 V80 Ω=0.6 AI = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{48\text{ V}}{80\text{ }\Omega} = 0.6\text{ A}
Current is uniform throughout all components connected in a series loop.
3
Calculate the power dissipation in the load resistor using Watt's Law
P=I2×R=(0.6 A)2×50 Ω=0.36 A2×50 Ω=18 WP = I^2 \times R = (0.6\text{ A})^2 \times 50\text{ }\Omega = 0.36\text{ A}^2 \times 50\text{ }\Omega = 18\text{ W}
Power dissipated by a resistor is proportional to the square of the current flowing through it multiplied by its resistance.

Key Concept

Series circuit current and resistor power dissipation using combined Ohm's Law and Watt's Law (P=I2RP = I^2 R or P=V2RP = \frac{V^2}{R})

Alternative Method

Find the voltage drop across the load resistor using the voltage divider rule: Vload=48 V×(50 Ω30 Ω+50 Ω)=48×58=30 VV_{\text{load}} = 48\text{ V} \times \left(\frac{50\text{ }\Omega}{30\text{ }\Omega + 50\text{ }\Omega}\right) = 48 \times \frac{5}{8} = 30\text{ V}. Then compute power using P=Vload2Rload=30250=90050=18 WP = \frac{V_{\text{load}}^2}{R_{\text{load}}} = \frac{30^2}{50} = \frac{900}{50} = 18\text{ W}.
Estimated Time:1m 15s
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Electrical Power

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