In parallel circuits, the same voltage is applied to every branch, and the current from the source splits among those branches. The total current leaving the source is the sum of the currents through each branch, because at the junction where branches meet, currents recombine and Kirchhoff’s current law applies. Each branch carries a current I_i = V / R_i, since the branch voltage is V. Adding up all branch currents gives It = I1 + I2 + …. For a quick check, if you have 12 V across two branches with resistances of 6 Ω and 3 Ω, the branch currents are 2 A and 4 A, respectively, so the total current is 6 A. The equivalent resistance is 12 V / 6 A = 2 Ω, which also satisfies 1/R_eq = 1/6 + 1/3. The voltage across each branch is equal to the source voltage, not greater. The idea of dividing the source voltage by the sum of resistors applies to a single path, not to a parallel network, where the current distribution depends on each branch’s resistance.

Current is the same through every component in a series circuit. Since there’s only one path for the current, the same amount of charge that flows into the first component must flow through and out of the last, so the current you measure is the same everywhere along the loop. The voltage, however, can differ from one component to the next because V = I·R, and the current I is the same while each component may have a different resistance. The sum of all these voltages equals the total supply voltage. Remember, in a series circuit there isn’t a split of current, so you don’t add branch currents to get the total current. And the total resistance isn’t independent of component values; it’s the sum of the individual resistances, so changing any component changes the overall resistance.

In a parallel circuit, the voltage across every branch equals the source voltage. Since resistor three is directly across a 24-volt battery, the voltage across it must be 24 volts. The resistor values affect how much current each branch draws, not the voltage across the branches in parallel. For example, each 6-ohm resistor would carry I = 24/6 = 4 A, and the total current would be 16 A with a combined parallel resistance of 1.5 ohms, but those details don’t change the voltage across resistor three.

When capacitors are connected in parallel, they all share the same voltage, and the total storage comes from simply adding up what each one can hold at that voltage. For each capacitor, the charge it stores is Q = C V. With the same V across all capacitors, the total charge is Q_total = (C1 + C2 + ... ) V. Since total capacitance is defined by C_total = Q_total / V, it follows that C_total = C1 + C2 + ... . So the total capacitance is the sum of the individual capacitances. For example, a 4 μF and a 6 μF capacitor in parallel yield 10 μF. The other options would imply reductions or averaging, which don’t happen in parallel connections.

The current drawn from the bus depends on the input power, which equals the output power divided by the efficiency. Start by converting the motor’s output power to watts: 0.5 horsepower is 0.5 × 746 ≈ 373 watts. With 85% efficiency, the input power is P_in = 373 W / 0.85 ≈ 439 W. The bus voltage is 30 V, so the current is I = P_in / V ≈ 439 W / 30 V ≈ 14.6 A. So the motor draws about 14.6 amperes from the 30-volt bus. If you ignored efficiency, you’d get about 12.4 A, which would be incorrect for this case because not all input power becomes useful output due to losses.