Physics Practice Test: Electricity and Circuits

Ohm's law: I = V/R = 12 V ÷ 4.0 Ω = 3.0 A. Multiplying gives 48, and dividing R by V gives 0.33.

In series the resistances add: 2.0 + 4.0 + 6.0 = 12 Ω, so I = 24 ÷ 12 = 2.0 A. Using only the 6.0 Ω resistor gives 4.0 A.

1/R = 1/6.0 + 1/3.0 = 0.5, so R = 1 ÷ 0.5 = 2.0 Ω. Stopping at 0.50 forgets to take the reciprocal, and 9.0 Ω is the series total. A parallel combination is always smaller than its smallest resistor.

P = VI, so I = P/V = 60 W ÷ 120 V = 0.50 A. Dividing V by P gives 2.0, and multiplying gives 7200.

Energy = power × time = 1.5 kW × 4.0 h = 6.0 kWh. Keeping the power in watts gives 6000, which is in watt-hours, not kilowatt-hours; 1.5 kWh leaves out the time.

Two equal 8 Ω resistors in parallel give 8 ÷ 2 = 4 Ω; adding R1 in series gives 2 + 4 = 6.0 Ω. Adding all three as if in series gives 18 Ω.

The battery current is 12 V ÷ 6 Ω = 2.0 A. It splits equally between the two identical 8 Ω branches, so R2 carries 1.0 A. The value 2.0 A is the total current, and 1.5 A wrongly applies the full 12 V across R2.

R1 takes 2.0 A × 2 Ω = 4.0 V, leaving 12 − 4 = 8.0 V across the parallel part, and R2 and R3 each get that full 8.0 V. The value 4.0 V is the drop across R1.

Current is charge per time, so Q = It = 2.0 A × 30 s = 60 C. Dividing 30 by 2.0 gives 15 C, which treats current as time per charge.

Electrons are negative, so they are repelled by the negative terminal and drift toward the positive one. Conventional current is defined in the opposite direction, from positive to negative, which is the trap. Back-and-forth motion describes alternating current, not a battery circuit.

A series circuit has only one path, so a break anywhere stops the current everywhere. In a parallel circuit the other lamps would stay lit.

Coulomb's law, F = kq₁q₂/r², is an inverse-square law, so tripling r divides the force by 3² = 9. "One-third" ignores the square.

Resistance is R = ρL/A. Doubling L doubles R, and doubling A halves it, so the two changes cancel. Answering "doubles" accounts for the length but forgets the thicker wire.

P = I²R = 0.50² × 20 = 5.0 W. The value 10 is IR, which is the voltage across the resistor in volts, not the power.

Voltage is a difference between two points, so the voltmeter goes across the lamp, in parallel. Connecting in series is how an ammeter is used to measure current.