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Electrical Circuits and Ohm's Law Flashcards

6 cards from real BMST practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 6 Electrical Circuits and Ohm's Law flashcards as text

  1. A resistor network consists of three resistors in parallel: R1 = 6 Ω, R2 = 12 Ω, and R3 = 4 Ω. What is the total equivalent resistance of the network?

    Answer: 2 Ω

    For resistors in parallel, 1/R_total = 1/R1 + 1/R2 + 1/R3 = 1/6 + 1/12 + 1/4 = 2/12 + 1/12 + 3/12 = 6/12 = 1/2, so R_total = 2 Ω. A common mistake is adding the resistances directly (series formula) or averaging them.

  2. A 120 V source powers a circuit with two resistors in series: R1 = 30 Ω and R2 = 10 Ω. If R2 is replaced with a short circuit (0 Ω), what current flows through R1?

    Answer: 4 A

    When R2 is shorted, total resistance = R1 + 0 = 30 Ω. Current = V/R = 120/30 = 4 A. The original circuit had I = 120/40 = 3 A. Shorting R2 removes its resistance from the series path, increasing current through R1.

  3. An electric heater draws 8 A from a 240 V supply. Over 45 minutes of operation, how much electrical energy does it consume in kilowatt-hours?

    Answer: 1.44 kWh

    Power = V × I = 240 × 8 = 1,920 W = 1.92 kW. Time = 45 min = 0.75 hours. Energy = Power × Time = 1.92 × 0.75 = 1.44 kWh. A common error is forgetting to convert minutes to hours before multiplying.

  4. In a circuit, a 9 V battery with an internal resistance of 1 Ω is connected to an external resistor of 8 Ω. What is the terminal voltage (voltage across the external resistor) under load?

    Answer: 8 V

    Total resistance = 1 + 8 = 9 Ω. Current = 9V / 9Ω = 1 A. Voltage drop across internal resistance = 1 A × 1 Ω = 1 V. Terminal voltage = 9 V − 1 V = 8 V. The internal resistance causes a voltage drop that reduces the usable terminal voltage, a concept often overlooked in basic circuit analysis.

  5. Two resistors R1 = 20 Ω and R2 = 80 Ω are connected in parallel across a 40 V source. What is the ratio of power dissipated in R1 to power dissipated in R2?

    Answer: 4:1

    In a parallel circuit, voltage across each resistor is equal (40 V). Power = V²/R. P1 = 40²/20 = 1,600/20 = 80 W. P2 = 40²/80 = 1,600/80 = 20 W. Ratio P1:P2 = 80:20 = 4:1. Smaller resistance dissipates MORE power. This is counterintuitive because in a series circuit, larger resistance dissipates more power.

  6. A technician measures the voltage across a resistor as 15 V and the current through it as 250 mA. She then doubles the voltage across the resistor while keeping the resistor value unchanged. What is the new power dissipation?

    Answer: 15 W

    Original R = V/I = 15/0.25 = 60 Ω. Original power = 15 × 0.25 = 3.75 W. New voltage = 30 V. Because P = V²/R, doubling the voltage multiplies power by 4: 3.75 × 4 = 15 W. Verified: P = (30)²/60 = 900/60 = 15 W. A common mistake is assuming power only doubles when voltage doubles, forgetting the squared relationship.