RC series circuit (time-domain KVL): In a single-loop resistor–capacitor series circuit, is the instantaneous source voltage equal to the sum of the instantaneous drops across the capacitor and the resistor, i.e., v_s(t) = v_C(t) + v_R(t)?

Introduction / Context:
Kirchhoff's Voltage Law (KVL) states that the algebraic sum of voltages around any closed loop is zero at every instant. In a simple RC series loop, this means the applied source voltage must equal the sum of the instantaneous voltage drops across the resistor and the capacitor. The statement v_s(t) = v_C(t) + v_R(t) is therefore a direct application of KVL in the time domain.

Given Data / Assumptions:

• Single source, single-loop series circuit with components R and C.
• Ideal lumped components, no parasitic elements or mutual coupling.
• Voltages are instantaneous time-domain quantities (not merely magnitudes).

Concept / Approach:
KVL is a fundamental conservation law resulting from the conservative nature of electrostatic fields in lumped circuits. For a series RC, the same current i(t) flows through both elements. The resistor drop is v_R(t) = i(t) * R, while the capacitor drop is v_C(t) related to charge q(t) by v_C(t) = q(t) / C and i(t) = dq/dt. Regardless of the excitation (step, sinusoid, pulse), KVL holds instant by instant: v_s(t) = v_R(t) + v_C(t).

Step-by-Step Solution:

Verification / Alternative check:
For a DC step, v_C(t) rises from 0 toward V, v_R(t) decays from V toward 0, and their instantaneous sum remains V at all times, confirming KVL. For AC sinusoidal excitation, the phasor relation is Ṽs = ṼR + ṼC; note that magnitudes do not add arithmetically, but the complex phasors do.

Why Other Options Are Wrong:

Common Pitfalls:
Confusing instantaneous addition with RMS magnitude addition; in AC, phasor vector addition is required. Saying magnitudes add is wrong, but the time-domain sum is always valid.

Final Answer:
Correct