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

A resistor rated at 100 Ω ± 5% is connected to a 12 V source. What is the MAXIMUM possible current through the circuit?

Answer: 0.126 A

Maximum current occurs at minimum resistance. A 5% tolerance means the resistor could be as low as 95 Ω. Applying Ohm's Law: I = V/R = 12/95 ≈ 0.126 A. The nominal value gives 12/100 = 0.120 A, and the maximum resistance (105 Ω) gives only 0.114 A. Current and resistance are inversely related, so minimum resistance always yields maximum current.

Three resistors — 6 Ω, 12 Ω, and 4 Ω — are connected in parallel across a 24 V source. What is the total power dissipated by the circuit?

Answer: 288 W

Find the parallel equivalent: 1/R_total = 1/6 + 1/12 + 1/4 = 2/12 + 1/12 + 3/12 = 6/12, giving R_total = 2 Ω. Total power: P = V²/R = 576/2 = 288 W. Verify by summing branch powers — P₁=96 W, P₂=48 W, P₃=144 W — which also total 288 W. A common error is mistakenly adding resistances in series (R=22 Ω), which produces a wildly incorrect 26 W.

A 60 W bulb and a 100 W bulb (both rated at 120 V) are wired in SERIES to a 120 V source. Which bulb glows brighter?

Answer: The 60 W bulb, because it has higher resistance and dissipates more power in series

Each bulb's resistance at rated conditions: R = V²/P. So R₆₀ = 14,400/60 = 240 Ω and R₁₀₀ = 14,400/100 = 144 Ω. In series the same current flows through both. Power = I²R, so the higher-resistance 60 W bulb dissipates more power and glows brighter. This is counterintuitive — the lower-wattage bulb actually receives more power in a series connection because its resistance is higher.

R₁ = 15 Ω is in series with two parallel resistors R₂ = R₃ = 20 Ω, all powered by a 36 V source. What is the voltage drop across R₁?

Answer: 21.6 V

The parallel pair (two equal 20 Ω resistors) has an equivalent resistance of 20/2 = 10 Ω. Total circuit resistance = 15 + 10 = 25 Ω. Circuit current: I = 36/25 = 1.44 A. Voltage across R₁: V = 1.44 × 15 = 21.6 V. The remaining 14.4 V appears across the parallel section. Using the voltage divider shortcut: V_R1 = 36 × (15/25) = 21.6 V confirms the result.

A wire measures 0.5 Ω at 20°C. Its temperature coefficient of resistance is 0.004/°C. What is the resistance after the wire heats to 70°C during operation?

Answer: 0.60 Ω

Using the formula R_T = R₀ × [1 + α(ΔT)]: ΔT = 70 − 20 = 50°C. R_T = 0.5 × [1 + 0.004 × 50] = 0.5 × 1.20 = 0.60 Ω. This 20% resistance increase means 20% less current at the same voltage — critical knowledge for wire sizing and thermal derating in real installations.

A 9 V battery with 1 Ω internal resistance drives an 8 Ω external load. What percentage of the battery's total generated power is lost to internal resistance?

Answer: 11.1%

Total resistance = 1 + 8 = 9 Ω. Current: I = 9/9 = 1 A. Total power from the EMF source: P_total = EMF × I = 9 W. Power wasted internally: P_internal = I² × r = 1² × 1 = 1 W. Percentage lost = 1/9 × 100 ≈ 11.1%. The general rule is that the fraction of power lost equals r/(r + R_load) — showing why minimizing internal resistance is vital for battery efficiency and terminal voltage under load.