Circuit Theory 1 · Thevenin and Norton

#25 Thevenin and Norton #25 — power in a 4 Ω load

Reduces the independent-source network to Thevenin and Norton forms, then finds the 4 Ω load current, its actual direction, and absorbed power.

Question

Original circuit with a 4 Ω load between a-b and independent 12 V and 3 A sources. — The power absorbed by the 4 Ω load will be found from Thevenin and Norton equivalents.

Treat the 4 Ω resistor between a-b as the load. Find the remaining network's Thevenin and Norton equivalents, the load current, and the absorbed power.

Written solution and narration transcript(shows the full solution)

Below are all the lines written in the notebook together with the full narration transcript.

1. Problem
The power absorbed by the 4 Ω load will be found from Thevenin and Norton equivalents.
Target: power in the 4 Ω load
Load: 4 Ω between a-b
Remaining network → Thevenin equivalent
Remove the load first
Find V_Th and R_Th
Reconnect the 4 Ω load
Required: P_4Ω
Narration transcript
In this lesson we use Thevenin's theorem on a circuit with only independent sources. The target is the power absorbed by the 4 ohm resistor between terminals a and b. Instead of solving the full circuit directly, we treat that 4 ohm resistor as the load, remove it, and replace the rest of the network by its Thevenin equivalent.
2. Remove the load
The voltage reference is V_Th=V_a−V_b.
Remove the load
Take the 4 Ω resistor out
a-b become open terminals
$V_{\mathrm{T}}h=V_{\mathrm{a}}-V_{\mathrm{b}}$
Thevenin: V_Th in series with R_Th
Reconnect 4 Ω afterward
Keep the a→b reference
Narration transcript
The first move is to remove the 4 ohm load. The remaining network is seen from terminals a and b. Everything to the left and right of those terminals will become a voltage source V Thevenin in series with a resistance R Thevenin. After that, the 4 ohm resistor can be connected back to the equivalent.
3. R_Th
R_Th=(6∥12)+6=10 Ω.
R_Th: turn off sources
12 V → short circuit
3 A → open circuit
$6 \Omega ∥ 12 \Omega=4 \Omega$
This result is in series with 6 Ω
$R_{\mathrm{T}}h=4+6$
$R_{\mathrm{T}}h=10 \Omega$
Narration transcript
To find R Thevenin, turn off the independent sources. The 12 volt source becomes a short circuit, and the 3 amp current source becomes an open circuit. Looking between a and b, the left side is 6 ohms in parallel with 12 ohms. That parallel part is 4 ohms. From b to the bottom node there is another 6 ohm resistor in series with that result. Therefore R Thevenin is 4 plus 6, equal to 10 ohms.
4. V_Th
V_a=4 V, V_b=18 V, and V_Th=−14 V.
$V_{\mathrm{T}}h=V_{\mathrm{a}}-V_{\mathrm{b}}$
$V_{\mathrm{a}}/6+(V_{\mathrm{a}}-12)/12=0$
$3V_{\mathrm{a}}-12=0 \Rightarrow V_{\mathrm{a}}=4 V$
$V_{\mathrm{b}}/6-3=0$
$V_{\mathrm{b}}=18 V$
$V_{\mathrm{T}}h=4-18$
$V_{\mathrm{T}}h=-14 V$
Narration transcript
Now find the open-circuit voltage V Thevenin, equal to V a minus V b. For node a, KCL gives V a over 6 plus V a minus 12 over 12 equals zero. Multiplying by 12 gives 3 V a minus 12 equals zero, so V a is 4 volts. For node b, the 3 amp source injects current upward into the node, so V b over 6 minus 3 equals zero. Therefore V b is 18 volts. So V Thevenin is V a minus V b, which is 4 minus 18, equal to negative 14 volts.
5. Load current and power
With the a→b reference i=−1 A; actual current is b→a and P=4 W.
Reconnect the 4 Ω load
$i_{a\to b}=V_{\mathrm{T}}h/(R_{\mathrm{T}}h+4)$
$i_{a\to b}=-14/(10+4)$
$i_{a\to b}=-1 A$
Actual direction: b→a
$P=i^{2}R$
$P_{4\Omega}=1^{2}\cdot 4=4 W$
Narration transcript
Reconnect the 4 ohm resistor to the Thevenin equivalent. Using the current reference from a to b, the load current is V Thevenin divided by R Thevenin plus 4 ohms. That is negative 14 over 14, equal to negative 1 ampere. The negative sign only says that the actual current flows from b to a. The absorbed power is i squared R, so one amp squared times 4 ohms gives 4 watts.
6. Norton check
The Norton form also gives 4 W in the load.
Check with Norton
$I_{\mathrm{N}}=V_{\mathrm{T}}h/R_{\mathrm{T}}h$
$I_{\mathrm{N}}=-14/10$
$I_{\mathrm{N}}=-1.4 A (a\to b)$
$R_{\mathrm{N}}=R_{\mathrm{T}}h=10 \Omega$
10 Ω is parallel with the source
The load still absorbs 4 W
Narration transcript
The same result can be checked with Norton form. I Norton equals V Thevenin divided by R Thevenin, so it is negative 14 over 10, equal to negative 1.4 amperes for the a to b reference direction. The resistance is still 10 ohms in parallel. Whether we use Thevenin or Norton, the load terminals see the same behavior and the 4 ohm resistor absorbs 4 watts.
7. Method summary
With the a→b reference i=−1 A; actual current is b→a and P=4 W.
Results
$R_{\mathrm{T}}h=10 \Omega$
$V_{\mathrm{a}}=4 V, V_{\mathrm{b}}=18 V$
$V_{\mathrm{T}}h=-14 V$
$i_{a\to b}=-1 A$
Actual current flows b→a
$P_{4\Omega}=4 W$
Narration transcript
The key details are the terminal reference and the source deactivation rules. Here, the load is the 4 ohm resistor between a and b. Turning off the independent sources gives R Thevenin equal to 10 ohms. The open circuit node voltages are V a equal to 4 volts and V b equal to 18 volts, so V Thevenin is negative 14 volts. After reconnecting the 4 ohm load, the current reference is negative 1 ampere and the absorbed power is 4 watts.