AC voltages and currents vary with time, so we use a single number that reflects the actual power they can deliver. That number is the RMS (root-mean-square) or effective value. It’s defined so that the power in a resistor with AC voltage equals the same as with a steady DC voltage of the same magnitude, i.e., P = V_rms^2 / R = I_rms^2 R. For a sinusoidal signal, V_rms is the peak voltage divided by sqrt(2); the same relation holds for current. If we used peak values, power ratings wouldn’t have a consistent basis; the average value over a full cycle for a sine wave is zero, which isn’t useful for power. Instantaneous values describe the exact momentary reading, but ratings and comparisons require a stable measure, hence RMS. Devices and systems are thus specified by effective (RMS) values.

When the landing gear are down, the warning horn system is designed to stay quiet unless there’s a true fault. Placing the control valve switch in the neutral position isolates the throttle-position input from the horn circuit, so closing the throttles won’t trigger the horn. This keeps the cockpit from hearing a nuisance warning during approach or after landing when gear are down. Therefore, the correct outcome is to prevent the warning horn from sounding when the throttles are closed. The other options would either imply delaying or triggering the horn, or having no effect, which doesn’t align with how the interlock is intended to work.

Reading a 5-band resistor involves three significant digits, a multiplier, and a tolerance. Brown, Black, Black give 1-0-0, so the digits are 100. The multiplier band is red, which represents 10^2 (100), so 100 × 100 = 10,000 ohms. The final band is gold, meaning a tolerance of ±5%. So the resistor is 10 kΩ with ±5% tolerance.

In a parallel circuit, the total resistance depends on how many separate paths for current exist. Each extra path lowers the overall resistance because current has more routes to flow through. For three identical bulbs in parallel, the combined resistance is one-third of a single bulb’s resistance. If one bulb is removed, only two paths remain, so the total resistance becomes one-half of a single bulb’s resistance, which is higher than the three-path case. So the total resistance increases. It wouldn’t decrease, stay the same, or become zero as long as at least one path remains.