Get ready for the FCC General Radiotelephone Operator License (GROL) exam. Study with flashcards and multiple choice questions, each with hints and detailed explanations. Prepare effectively for your licensing test!

Multiple Choice

What is the voltage across the lamp in the circuit described in the first question?

Voltage across a lamp in a simple circuit comes from how the total supply voltage is shared among the series elements. The lamp’s voltage is the current times the lamp’s resistance, Vlamp = I × Rlamp, and the current is set by the total resistance in the loop: I = Vs / (Rlamp + Rseries). So if there’s a small extra resistance in series, the lamp won’t get the full supply; a tiny portion of the source voltage drops across that other element. In the described circuit, the source is about 6 volts, and there is a small series drop. When you work through the numbers with the given values, the lamp ends up with 5.994 volts across it. That’s just about 6 volts minus a small amount (the drop across the other series resistance), which makes sense for a small series resistance relative to the lamp’s resistance. The other nearby values would imply larger drops elsewhere or exact full supply across the lamp, which is not consistent with the described setup. So the 5.994 V result is the correct one.

Voltage across a lamp in a simple circuit comes from how the total supply voltage is shared among the series elements. The lamp’s voltage is the current times the lamp’s resistance, Vlamp = I × Rlamp, and the current is set by the total resistance in the loop: I = Vs / (Rlamp + Rseries). So if there’s a small extra resistance in series, the lamp won’t get the full supply; a tiny portion of the source voltage drops across that other element.

In the described circuit, the source is about 6 volts, and there is a small series drop. When you work through the numbers with the given values, the lamp ends up with 5.994 volts across it. That’s just about 6 volts minus a small amount (the drop across the other series resistance), which makes sense for a small series resistance relative to the lamp’s resistance. The other nearby values would imply larger drops elsewhere or exact full supply across the lamp, which is not consistent with the described setup. So the 5.994 V result is the correct one.