Tech Note D

When a load is not the only thing attached

The load formulas assume a load and nothing else. When something more is connected to the equivalent, Symbulator will write the equivalent circuit into the input fields for you, ready to be added to.

Last updated 2026-09-11

Lesson 4's load answers — irl, vrl, prl and pmax — are formulas, and they hold only when a load is the one thing connected to the equivalent circuit. Connect anything else to those terminals and none of them applies any more, and the problem has to be run as a circuit again.

Symbulator will write that circuit for you, so you are adding to it rather than typing it out.

The button

With the load question ticked, a button appears at the foot of Results, under the equivalent's answers: Load circuit equivalent? It is live once the equivalent has been found, and goes dead again if you change any input until you run again.

Press it and it warns you first: the equivalent circuit will overwrite the Circuit Description, Define and Expert Mode fields and switch the analysis to Solve circuit, so a circuit you have not saved yet can be saved before it goes. Proceed, and the description becomes

Circuit Description

jN,0,n,iNo
rE,n,0,rEq
rL,n,0,load

with Define holding the values of iNo and rEq that Results found, exact rather than rounded, and a value for load if you had given it one. It is the Norton equivalent connected, between nodes n and 0, to a load called rL with the symbolic value load, in ohms. You can use it as a starting point.

An example

RM3's Example 9-8

Open in app ↗Open in split view

Find the Norton equivalent of the circuit left of a-b; then find the current through R_L.

Solution

Let's first find the circuit equivalent:

Circuit Description

e,1,0,24
r1,1,2,120
r2,2,0,280
j,2,0,560'm

Find equivalent, Thévenin / Norton, nodes 2 and 0, in DC, with the load question ticked. Results gives:

Norton current

\[i_{no} = -0.36\ \mathrm{A}\]

equivalent resistance

\[R_{eq} = 84\ \Omega\]

The Norton current is reported in the direction it actually flows, from the first node to the second. Carry the sign through and the load current below comes out the same.

Now to the second part of the question. In order to find the current through R_L, we cannot use the load expressions, because now the load is not the only thing connected to the terminals of the equivalent: there is also a current source. We have to run a new simulation.

The fastest way is the button above. Proceed past its warning and the description is replaced by the equivalent, with iNo and rEq already in Define:

Circuit Description

jN,0,n,iNo
rE,n,0,rEq
rL,n,0,load

We change the value of the load to 168 Ω, and add the 180 mA source flowing from node 0 to node n. Then we run a dc simulation and ask for the current in the load:

Circuit Description

jN,0,n,iNo
rE,n,0,rEq
rL,n,0,168
j,0,n,180'm

with the loadOpen in app ↗Open in split view

The analysis is already Solve circuit; run it in DC, with SI prefixes ticked in Settings. The current through rl, irl, reads -60 mA.

Correct: there is a current of 60 mA flowing through R_L from 0 to n.

Using the equivalent circuit description is meant to save you time. If you find it confusing to use, just don't use it.