- Basic Math and Science Electrical Circuits and Ohm's Law Questions and Answers Flashcards

A resistor network consists of three resistors in parallel: 6 Ω, 12 Ω, and 4 Ω. What is the equivalent resistance of this combination?

Answer: 2 Ω

For parallel resistors: 1/R_eq = 1/6 + 1/12 + 1/4. Finding a common denominator of 12: 2/12 + 1/12 + 3/12 = 6/12 = 1/2, so R_eq = 2 Ω. A common mistake is simply adding the values (6+12+4 = 22 Ω), which only applies to series circuits. Adding parallel resistors always produces a result smaller than the smallest individual resistor.

A 120 V source drives a series circuit containing a 10 Ω resistor and an unknown resistor R₂. If the voltage drop across the 10 Ω resistor is 40 V, what is the value of R₂?

Answer: 20 Ω

First find the current using the known resistor: I = V/R = 40V / 10Ω = 4 A. In a series circuit this same 4 A flows through every component. The remaining voltage across R₂ = 120V − 40V = 80V. Therefore R₂ = V/I = 80V / 4A = 20 Ω. The distractor 30 Ω results from incorrectly using total voltage (120V / 4A), and 80 Ω comes from forgetting to divide the remaining voltage by current.

A 12 Ω and a 6 Ω resistor are connected in parallel; this combination is then wired in series with a 4 Ω resistor across a 48 V supply. What is the power dissipated by the 4 Ω series resistor?

Answer: 144 W

Step 1 — parallel combination: 1/R_p = 1/12 + 1/6 = 1/12 + 2/12 = 3/12, so R_p = 4 Ω. Step 2 — total series resistance: R_total = 4 + 4 = 8 Ω. Step 3 — total current: I = 48V / 8Ω = 6 A. Step 4 — power in 4 Ω: P = I²R = (6)² × 4 = 36 × 4 = 144 W. The distractor 72 W results from using P = V×I with only half the supply voltage, and 288 W from incorrectly doubling.

A technician measures 0.5 A flowing through a 200 Ω resistor rated for a maximum of 10 W. Is the resistor operating safely, and what is its actual power dissipation?

Answer: Unsafe; 50 W dissipated

Power = I²R = (0.5)² × 200 = 0.25 × 200 = 50 W. Since 50 W far exceeds the 10 W maximum rating, the resistor is operating unsafely and risks overheating or failure. The distractor '100 W' incorrectly uses I = 1 A, and '5 W' uses I² = 0.025 instead of 0.25. Marking a component safe when it dissipates 5× its rated power is a critical error in circuit analysis.

Three identical light bulbs (each with resistance R) are connected in series across a 90 V source. They are then reconnected in parallel across the same 90 V source. How does the total current drawn from the source in the parallel arrangement compare to the series arrangement?

Answer: 9 times greater

Series: R_total = 3R, so I_series = 90V / 3R = 30/R. Parallel: each bulb sees the full 90 V, drawing 90/R each; total I_parallel = 3 × (90/R) = 270/R. Ratio = (270/R) ÷ (30/R) = 9. The source must supply 9 times more current in parallel. This is why overloaded parallel circuits blow fuses — each added branch multiplies current demand, not just adds to it proportionally.

A 25-meter wire with resistance of 0.02 Ω per meter is connected in series with a 3.5 Ω load across a 12 V battery. What percentage of the supply voltage is lost across the wire?

Answer: 12.5%

Wire resistance = 0.02 Ω/m × 25 m = 0.5 Ω. Total circuit resistance = 0.5 + 3.5 = 4.0 Ω. Current I = 12V / 4Ω = 3 A. Voltage drop across wire = 3A × 0.5Ω = 1.5 V. Percentage loss = (1.5V / 12V) × 100 = 12.5%. This wiring loss means the load only receives 10.5 V instead of 12 V — a significant derating that matters in automotive and low-voltage installations.