Switched Two-Source AC Loop: 04-BS-4, December 2017, Q4

Question 4: Switched Two-Source AC Loop (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Left branch $R$-$L$ (top node marked $v_1(t)$) in series with source $v_{s1}(t)$ up to the switch node; right branch $v_{s2}(t)$ in series with $R_{LOAD}$; switch $S$ ties the mid node to the common bottom rail (ground).

Given data
$R$	$L$	$R_{LOAD}$	$v_{s1}(t)$	$v_{s2}(t)$	$\omega$
10Ω	1 mH	100Ω	20cos(120πt+π/3) V	10cos(120πt) V	120π rad/s

Find. (a) $I_1,V_1$ with S closed; (b) $P_{R_{LOAD}}$ with S closed; (c) $I_1(t),I_2(t),V_1(t)$ with S open; (d) the change in $P_{R_{LOAD}}$.

Figure 4: Two-source loop switched by S

Approach. With S closed the mid node is grounded, splitting the circuit into two independent single-source loops. With S open, the mid node floats and the whole network becomes one series loop carrying a single current driven by both sources together.

a) S closed — left loop. Grounding the mid node makes $v_{s1}$ drive $R+j\omega L$ alone (ground → $L$ → $R$ → node → $v_{s1}$ → ground). With $Z_L=j(120\pi)(0.001)=j0.377\,\Omega$: $$I_1=\frac{V_{s1}}{R+j\omega L}=\frac{20\angle 60^\circ}{10+j0.377}=\boxed{1.999\angle 57.84^\circ\text{ A}}$$ $$V_1 = -I_1(R+j\omega L) = \boxed{20\angle{-120^\circ}\text{ V}}$$ (By KVL around the single loop $V_1$ is simply $-V_{s1}$, independent of $R,L$; $I_1$ is what those elements draw to satisfy it.)
b) S closed — power in $R_{LOAD}$. Similarly, the right loop is driven by $v_{s2}$ alone: $I_2=V_{s2}/R_{LOAD}=0.1\angle 0^\circ$ A. $$P_{R_{LOAD}}=\tfrac12|I_2|^2R_{LOAD}=\tfrac12(0.1)^2(100)=\boxed{0.500\text{ W}}$$
c) S open — single series loop. The mid node no longer grounds; $R,L,v_{s1},v_{s2},R_{LOAD}$ now form ONE loop, so $I_1(t)=I_2(t)$ (same current, confirming the labelling): $$I=\frac{V_{s1}+V_{s2}}{R+j\omega L+R_{LOAD}}=\frac{20\angle60^\circ+10\angle0^\circ}{110+j0.377}=\boxed{0.2405\angle 40.70^\circ\text{ A}} = I_1(t)=I_2(t)$$ $$V_1 = -I(R+j\omega L) = \boxed{2.407\angle{-137.14^\circ}\text{ V}}$$
d) Change in load power. $$P_{R_{LOAD},open}=\tfrac12|I|^2R_{LOAD}=\tfrac12(0.2405)^2(100)=2.893\text{ W}$$ $$\Delta P = P_{open}-P_{closed}=2.893-0.500=\boxed{+2.39\text{ W (increase)}}$$

Final results — Question 4
Quantity	Value
$I_1$ (S closed)	1.999∠57.84° A
$V_1$ (S closed)	20∠−120° V
$P_{R_{LOAD}}$ (S closed)	0.500 W
$I_1=I_2$ (S open)	0.2405∠40.70° A
$V_1$ (S open)	2.407∠−137.14° V
$P_{R_{LOAD}}$ (S open)	2.893 W
ΔP	+2.39 W

Check: reading the source polarity marks (+ terminal toward the switch for $v_{s1}$, + terminal toward $R_{LOAD}$ for $v_{s2}$) from the printed figure was essential — a flipped reference would change every phase angle above.

Topic: Steady-state phasor analysis of a switch-reconfigured AC network.

Key relations: single-loop KVL $\sum V_{sources}=I\sum Z$; grounding a node with an ideal switch decouples the branches on either side into independent loops.

Why this works: closing S ties the mid node to the reference potential, so each side's source only "sees" its own local impedance; opening S removes that reference tie, forcing a single shared current path and reactivating the algebraic sum of both sources as the net driving phasor.

Common pitfall: assuming $I_1\ne I_2$ when S is open — in a single series loop every element carries the same current, which the exam's own separate $I_1(t)$/$I_2(t)$ request is implicitly testing for.

Source: Sadiku & Alexander, Fundamentals of Electric Circuits, Ch. 9–10 (phasors, AC steady-state).

Question 4 of 7: Switched Two-Source AC Loop

Question 4: Switched Two-Source AC Loop (20 marks)