Examkin

Question

Difficulty: HardElectrical Power

A continuous-duty DC servomotor coil is connected to a 50 V50\text{ V} power line and draws a steady current of 4 A4\text{ A}. During a brownout condition, the source voltage drops to 40 V40\text{ V} while the internal resistance of the coil remains constant. What is the electrical power dissipated by the servomotor coil under this reduced voltage condition?

  1. A
    40 W40\text{ W}
  2. 128 W128\text{ W}Answer
  3. C
    160 W160\text{ W}
  4. D
    200 W200\text{ W}

Answer

128 W128\text{ W}
The coil's resistance is determined first using Ohm's Law: R=50 V4 A=12.5 ΩR = \frac{50\text{ V}}{4\text{ A}} = 12.5\text{ }\Omega. Under the reduced supply voltage of 40 V40\text{ V}, power dissipation is calculated using P=V2R=40212.5=160012.5=128 WP = \frac{V^2}{R} = \frac{40^2}{12.5} = \frac{1600}{12.5} = 128\text{ W}. Alternatively, the new current is I=40 V12.5 Ω=3.2 AI = \frac{40\text{ V}}{12.5\text{ }\Omega} = 3.2\text{ A}, giving P=40 V×3.2 A=128 WP = 40\text{ V} \times 3.2\text{ A} = 128\text{ W}.

Step-by-Step Solution

1
Calculate the constant internal resistance of the coil from initial operating values
R=V1I1=50 V4 A=12.5 ΩR = \frac{V_1}{I_1} = \frac{50\text{ V}}{4\text{ A}} = 12.5\text{ }\Omega
The coil's physical resistance remains constant despite changes in external supply voltage.
2
Calculate the new current drawn by the coil at the reduced voltage of 40 V40\text{ V}
I2=V2R=40 V12.5 Ω=3.2 AI_2 = \frac{V_2}{R} = \frac{40\text{ V}}{12.5\text{ }\Omega} = 3.2\text{ A}
By Ohm's Law, reduced voltage across a fixed resistance causes a proportional reduction in current.
3
Calculate the new electrical power dissipated under the brownout condition
P2=V2×I2=40 V×3.2 A=128 WP_2 = V_2 \times I_2 = 40\text{ V} \times 3.2\text{ A} = 128\text{ W} (or P2=V22R=40212.5=128 WP_2 = \frac{V_2^2}{R} = \frac{40^2}{12.5} = 128\text{ W})
Watt's Law defines power as the product of voltage and current or the square of voltage divided by resistance.

Key Concept

Electrical power in a resistive load varies with the square of voltage (P=V2RP = \frac{V^2}{R}) when resistance is held constant.

Alternative Method

Proportional scaling using Watt's Law: Since P∝V2P \propto V^2 for constant resistance, P2=P1×(V2V1)2=200 W×(4050)2=200×0.64=128 WP_2 = P_1 \times \left(\frac{V_2}{V_1}\right)^2 = 200\text{ W} \times \left(\frac{40}{50}\right)^2 = 200 \times 0.64 = 128\text{ W}.
Estimated Time:1m 30s
Rate this question

Topics

Electrical Power

More questions from this topic