In a series RLC circuit, the branch magnitudes are V_R = I_T * R, V_L across the inductor, and V_C across the capacitor. Given I_T = 3 mA, V_L = 30 V, V_C = 18 V, and R = 1000 Ω, what is the magnitude of the applied source voltage?

Introduction:
In series RLC circuits, branch voltages are not added arithmetically because they are out of phase. The resistor voltage V_R is in phase with the current, while V_L leads and V_C lags the current by 90 degrees, making V_L and V_C 180 degrees apart from each other. The source voltage is the phasor (vector) sum of these components.

Given Data / Assumptions:

• Total current: I_T = 3 mA.
• Inductor voltage: V_L = 30 V (leads I by +90°).
• Capacitor voltage: V_C = 18 V (lags I by −90°).
• Resistance: R = 1000 Ω ⇒ V_R = I_T * R.
• Sinusoidal steady state, ideal elements.

Concept / Approach:
Compute V_R = I_T * R. Reactive voltages along j-axis combine algebraically: V_X = V_L − V_C. The applied voltage magnitude is |V_S| = sqrt(V_R^2 + V_X^2). This uses orthogonal decomposition of phasors (real axis for V_R, imaginary axis for net reactance voltage).

Step-by-Step Solution:
V_R = I_T * R = 0.003 A * 1000 Ω = 3.0 VNet reactive voltage: V_X = V_L − V_C = 30 − 18 = 12 VApplied voltage magnitude: |V_S| = sqrt(V_R^2 + V_X^2)|V_S| = sqrt(3.0^2 + 12^2) = sqrt(9 + 144) = sqrt(153) ≈ 12.369… VRounded to two decimals: 12.37 V

Verification / Alternative check:
Phasor diagram: V_R on real axis = 3 V; V_L and V_C on imaginary axis, opposing. Resultant vector length matches the computed magnitude. The value is sensible because the reactive voltages largely cancel, leaving a moderate resultant dominated by the 12 V net reactance and a small 3 V resistive component.

Why Other Options Are Wrong:

• 3.00 V: Ignores the net reactive contribution (12 V).
• 34.98 V / 48.00 V: Resemble arithmetic sums or partial sums of branch magnitudes; phasors must be combined vectorially.
• 9.00 V: Equal to V_R + (V_L − V_C) linearly; again, incorrect because of orthogonal components.

Common Pitfalls:

• Adding V_R, V_L, and V_C directly without phasor resolution.
• Forgetting that V_L and V_C oppose each other by 180 degrees along the reactive axis.
• Rounding too early; keep sufficient precision until the final step.

Final Answer:
12.37 V