Exercises — Apply superposition theorem

• Ideal voltage source → short circuit (a plain wire). Killing it means "force $0\text{ V}$ across its terminals". Two terminals at the same potential with a wire between them = $0\text{ V}$. A wire allows any current, exactly what a $0\text{ V}$ source does.
• Ideal current source → open circuit (a gap). Killing it means "force $0\text{ A}$ through it". No path = no current, which is a gap.
• Remember the mnemonic "V-Short, I-Open".

1. Dependent sources are never deactivated. They are controlled by circuit variables (a voltage or current elsewhere), so removing one removes real physics. Only independent sources get toggled.
2. Power does not superpose. Power is $P=i^{2}R$, which contains a square — a nonlinear operation. Superposition only holds for the linear quantities (voltages, currents). Find total current/voltage first, then compute power once.

Killing the current source means opening it (a gap). Now $R_1$ and $R_2$ form a single series path from $12\text{ V}$ to ground, with $V_A$ tapped between them — a plain voltage divider: $V_{A,1}=12\cdot \frac{R_2}{R_1+R_2}=12\cdot \frac{6}{4+6}=12\cdot 0.6=7.2\text{ V}$ Why a divider? With the gap in place, all the current from the source flows through $R_1$ then $R_2$; the fraction of the $12\text{ V}$ that lands across $R_2$ is exactly $R_2/(R_1+R_2)$.

Killing the voltage source means shorting it (a wire). Now the top of $R_1$ is connected straight to ground, so $R_1$ and $R_2$ both run from node $A$ to ground — they are in parallel: $R_1\Vert R_2=\frac{R_1{R}_2}{R_1+R_2}=\frac{4\cdot 6}{4+6}=\frac{24}{10}=2.4\Omega$ All $3\text{ A}$ leaves node $A$ through this parallel resistance to ground, so $V_{A,2}=I\cdot (R_1\Vert R_2)=3\times 2.4=7.2\text{ V}$

Both contributions raise node $A$ above ground (same polarity), so they add directly: $V_A=V_{A,1}+V_{A,2}=7.2+7.2=14.4\text{ V}$

Fix the reference direction as downward (node $A$ → ground) and keep it for both sub-problems.

• From $12\text{ V}$ alone: $I_1=\frac{V_{A,1}}{R_2}=\frac{7.2}{6}=1.2\text{ A}$ (downward).
• From $3\text{ A}$ alone: $I_2=\frac{V_{A,2}}{R_2}=\frac{7.2}{6}=1.2\text{ A}$ (downward). Both point downward, so add: $I_{R_2}=1.2+1.2=2.4\text{ A}(\text{downward})$ Cross-check: $V_A/R_2=14.4/6=2.4\text{ A}$ ✓.

KCL: sum of currents leaving node $A$ equals zero. Current toward the $12\text{ V}$ source through $R_1$ is $(V_A-12)/4$; current to ground through $R_2$ is $V_A/6$; the $3\text{ A}$ source injects current, so it leaves as $-3$: $\frac{V_A-12}{4}+\frac{V_A}{6}-3=0$ Multiply through by $12$: $3(V_A-12)+2V_A-36=0⟹5V_A-72=0⟹V_A=14.4\text{ V}$ Matches the superposition answer, as linearity guarantees.

The voltage-source contribution is unchanged: $V_{A,1}=+7.2\text{ V}$. The current-source contribution now has the opposite polarity: current is drawn out of node $A$, so it lowers $V_A$: $V_{A,2}=-3\times 2.4=-7.2\text{ V}$ Add algebraically: $V_A=7.2+(-7.2)=0\text{ V}$ Node $A$ sits exactly at ground — the sources cancel. This is the whole point of algebraic (signed) addition.

Correct method — total current first, then power once: $I_{R_2}=2.4\text{ A}⟹P_{R_2}=I_{R_2}^2{R}_2=(2.4{)}^2\times 6=5.76\times 6=34.56\text{ W}$ The illegal shortcut would add per-source powers: $P_1=(1.2{)}^2\times 6=8.64\text{ W},P_2=(1.2{)}^2\times 6=8.64\text{ W},P_1+P_2=17.28\text{ W}$ That gives $17.28\text{ W}$, which is wrong ($\ne 34.56\text{ W}$). The reason: $(i_1+i_2{)}^2=i_1^2+2i_1{i}_2+i_2^2\ne i_1^2+i_2^2$ The missing $2i_1{i}_{2}R$ cross term is real energy. Here it equals $2(1.2)(1.2)(6)=17.28\text{ W}$, and indeed $17.28+17.28=34.56\text{ W}$ ✓. See Why power does not superpose.

With $R_2$ removed, node $A$ connects to: the $12\text{ V}$ source through $R_1=4\Omega$, and the $3\text{ A}$ source directly.

Open-circuit voltage $V_{th}$ (superposition, $R_2$ absent):

• $12\text{ V}$ alone (open the $3\text{ A}$): no current flows through $R_1$ into an open terminal, so no drop across $R_1$; the terminal sees the full $12\text{ V}$. $V_{oc,1}=12\text{ V}$.
• $3\text{ A}$ alone (short the $12\text{ V}$): all $3\text{ A}$ flows through $R_1$ to ground, raising node $A$ by $3\times 4=12\text{ V}$. $V_{oc,2}=12\text{ V}$. $V_{th}=12+12=24\text{ V}$ Thévenin resistance $R_{th}$: kill both independent sources (V→short, I→open). Only $R_1$ remains from node $A$ to ground: $R_{th}=R_1=4\Omega$ Reconnect $R_2=6\Omega$ — the equivalent is now a simple divider: $V_A=V_{th}\cdot \frac{R_2}{R_{th}+R_2}=24\cdot \frac{6}{4+6}=24\cdot 0.6=14.4\text{ V}✓$ Same $14.4\text{ V}$ — a totally independent confirmation.

We superpose three independent sources. Each time, kill the other two (V→short, I→open). Reference: $I_c$ downward ($A\to$ ground).

(a) $10\text{ V}$ alone (short $5\text{ V}$, open $2\text{ A}$): Node $A$ connects to ground through $R_a=2\Omega$ (leading to the $10\text{ V}$ source, now the only source), and through the parallel pair $R_b=5\Omega$ (its source shorted) and $R_c=10\Omega$. First find $V_A$ from a divider: $R_b\Vert R_c=\frac{5\cdot 10}{5+10}=\frac{50}{15}=3.333\Omega ,V_{A,1}=10\cdot \frac{R_b\Vert R_c}{R_a+R_b\Vert R_c}=10\cdot \frac{3.333}{2+3.333}=6.25\text{ V}$ $I_{c,1}=\frac{V_{A,1}}{R_c}=\frac{6.25}{10}=0.625\text{ A}$

(b) $5\text{ V}$ alone (short $10\text{ V}$, open $2\text{ A}$): By symmetry of method: $R_a\Vert R_c=\frac{2\cdot 10}{12}=1.667\Omega$, and $V_{A,2}=5\cdot \frac{R_a\Vert R_c}{R_b+R_a\Vert R_c}=5\cdot \frac{1.667}{5+1.667}=1.25\text{ V},I_{c,2}=\frac{1.25}{10}=0.125\text{ A}$

(c) $2\text{ A}$ alone (short both voltage sources): Now all three resistors run from node $A$ to ground in parallel: $R_a\Vert R_b\Vert R_c={\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)}^{-1}={\left(0.5+0.2+0.1\right)}^{-1}=\frac{1}{0.8}=1.25\Omega$ $V_{A,3}=2\times 1.25=2.5\text{ V},I_{c,3}=\frac{2.5}{10}=0.25\text{ A}$

Add algebraically (all three raise $V_A$, so all downward through $R_c$): $I_c=0.625+0.125+0.25=1.0\text{ A}$ Cross-check via total $V_A$: $V_A=6.25+1.25+2.5=10\text{ V}$, so $I_c=10/10=1.0\text{ A}$ ✓.

By linearity, $V_A$ is a sum of the fixed voltage-source part and the current-source part: $V_A=\underset{\text{from }12\text{ V}}{\underbrace{7.2}}+I_s\cdot (R_1\Vert R_2)=7.2+2.4I_s$ Set equal to $9\text{ V}$: $7.2+2.4I_s=9⟹2.4I_s=1.8⟹I_s=0.75\text{ A}$ A $0.75\text{ A}$ injection gives $V_A=9\text{ V}$. Notice how superposition turns the design task into one linear equation — no re-solving the whole circuit.