Series AC drops to peak source: Two resistors in series are connected to an AC source. If the measured voltages are 7.5 V rms across one resistor and 4.2 V rms across the other, what is the peak (maximum) source voltage?

Introduction / Context:
In basic AC circuit analysis with purely resistive series elements, the root-mean-square (rms) voltages across individual resistors add directly to give the total rms source voltage. Once the rms value is known, the peak value of a sinusoidal source can be found using the standard rms–peak relationship. This problem evaluates your fluency with these two ideas and correct unit handling.

Given Data / Assumptions:

• Two resistors are connected in series across a single sinusoidal AC source.
• Voltage drops: 7.5 V rms across one resistor and 4.2 V rms across the other.
• Assume ideal, purely resistive elements (voltages are in phase in series).
• Waveform is sinusoidal, so Vrms = Vpeak / √2.

Concept / Approach:
For series resistors in a purely resistive circuit, the same current flows through each and the phase is the same. Therefore, individual rms drops add arithmetically to obtain the total rms source voltage. Then use Vpeak = Vrms * √2 to convert to the peak (maximum) value of the sinusoid.

Step-by-Step Solution:

Verification / Alternative check:
Back-conversion: 16.54 V / 1.414 ≈ 11.7 V rms, matching the sum of the given drops, so the computation is self-consistent for a sinusoidal source with resistive series elements.

Why Other Options Are Wrong:

• 10.60 V: Close to the rms total, not the peak.
• 5.93 V and 1.65 V: These are far below the correct peak; they misuse the √2 factor or mix rms and peak values incorrectly.

Common Pitfalls:

• Adding peaks or mixing rms and peak magnitudes inconsistently.
• Using vector addition unnecessarily—here, resistive drops are in phase, so simple arithmetic addition is correct.

Final Answer:
16.54 V