Series RL vector addition: In a series RL circuit, 12 V (rms) is across the resistor and 14 V (rms) is across the inductor. What is the peak value of the source voltage?

Introduction / Context:This question evaluates AC phasor skills for series RL circuits. The resistor voltage is in phase with current, while the inductor voltage leads the current by 90°. The source voltage is the phasor (vector) sum of the two branch drops.

Given Data / Assumptions:

• VR = 12 V (rms) across R.
• VL = 14 V (rms) across L.
• Series RL, ideal components.
• Asked: Source voltage in peak (not rms).

Concept / Approach:For series RL: V_s(rms) = sqrt(VR^2 + VL^2). Convert that rms result to peak by multiplying by sqrt(2).

Step-by-Step Solution:V_s(rms) = sqrt(12^2 + 14^2)V_s(rms) = sqrt(144 + 196) = sqrt(340) ≈ 18.439 VV_s(peak) = V_s(rms) * sqrt(2)V_s(peak) ≈ 18.439 * 1.414 ≈ 26.08 V ≈ 26.0 V

Verification / Alternative check:Power triangle or impedance triangle gives the same phasor relationship; the orthogonal nature of VR and VL requires a Pythagorean sum for rms values.

Why Other Options Are Wrong:

• 18.4 V: That is the rms magnitude, not peak.
• 20 V, 2 V: Not consistent with the phasor sum and peak conversion.

Common Pitfalls:Adding VR and VL arithmetically, or forgetting to convert rms to peak. Always perform vector addition first, then scale to the requested unit (peak or rms).

Final Answer:26.0 V