Loading & equivalent resistance

Before RL is connected, the output node joins only R1 and R2, so one current flows through the series string. After RL is connected from VOUT to ground, the output node acquires a third branch. Current arriving through R1 can split between R2 and RL.

That means the original unloaded voltage-divider derivation no longer matches the new topology. The correct first step is to redraw or mentally reduce the actual network.

R2 and the load RL connect between the same two nodes, so they are parallel. Replace them by the lower equivalent resistance

The network is now R1 in series with Rlower. Therefore

This is not a new divider law. It is the ordinary voltage-divider result applied to the actual lower resistance after the load has been included.

Because R2∥RL is always smaller than R2 for finite positive RL, a passive resistive load connected to ground pulls this divider output below its unloaded value.

Take VIN=10 V and R1=R2=10 kΩ. Unloaded,

Now connect RL=10 kΩ. The lower resistance becomes

and the actual loaded output is

The relative drop from the unloaded value is

The failure is large because the load resistance is comparable with the divider resistances. The “5 V node” was never an ideal 5 V source.

A two-terminal equivalent resistance is a single resistance that reproduces the same voltage–current relation seen from a specified pair of terminals, within the model being used.

For a resistor-only linear network, series and parallel reductions are simply convenient ways of finding that terminal relation. The external circuit cannot distinguish the original passive network from Req if it interacts only through those two terminals.

But the replacement hides internal node voltages, individual branch currents, local power dissipation and component stress. If an internal node is used elsewhere, collapsing the network may destroy information that matters.

To ask how strongly the divider output reacts to a load, we need the resistance seen looking back into the output port. For a linear resistive circuit, independent ideal sources are first set to zero for this resistance calculation.

• An independent ideal voltage source set to 0 V becomes a short circuit.
• An independent ideal current source set to 0 A becomes an open circuit.
• Dependent sources are not simply turned off; the test-source definition must be used.

For the simple divider, setting VIN to 0 V shorts the input node to ground. R1 and R2 are then both connected from VOUT to ground:

For the 10 kΩ / 10 kΩ divider, Rout=5 kΩ.

After independent sources have been deactivated where appropriate, apply a test voltage Vt at the port and calculate the current It drawn into the network. Define

For two 10 kΩ resistors in parallel, a 1 V test source draws

so

This reproduces the parallel reduction but gives a definition that still works when visual series/parallel simplification is unavailable. In linear circuits containing dependent sources, the test source is especially important because those dependent sources remain active.

Let V0 be the unloaded output voltage and Rout the resistance seen from that output. For the linear resistive divider, direct algebra gives the loaded output

Define the fractional loading error as the relative drop from the unloaded output:

Substitution gives

If the design requires ε≤εmax, solve for the minimum load resistance:

For a 1% maximum loading error, the exact condition is

This is the quantitative origin of the common engineering preference for a load resistance roughly two orders of magnitude larger than the source/output resistance when very small DC loading error is desired.

A real voltmeter or digital multimeter (DMM) has finite input resistance. At DC, a simple first model is a resistor connected from the measured node to the instrument reference.

Case A — low-output-resistance divider. For the 10 kΩ / 10 kΩ divider, Rout=5 kΩ. With an illustrative 10 MΩ meter input,

The measurement loading is negligible for many purposes.

Case B — high-output-resistance source. If Rout=1 MΩ and the same 10 MΩ meter is used,

The meter now materially changes the quantity being measured.

The next conceptual step is powerful: a whole linear two-terminal source network can be represented by an ideal source plus one resistance as seen by a load. That is the Thévenin/Norton framework developed later in CT-01; DC-03 has now built every prerequisite needed to understand why it works.

1. A 12 V divider uses R1=20 kΩ and R2=10 kΩ. Find the unloaded output, then the loaded output with RL=10 kΩ.
2. For a 15 kΩ / 5 kΩ divider, find the resistance seen from the output after the ideal voltage source is set to zero.
3. A source has V0=2.5 V and Rout=2 kΩ. Find the loaded voltage for RL=18 kΩ and the relative loading error.
4. What minimum RL guarantees no more than 0.5% loading error if Rout=1 kΩ?
5. A 10 MΩ voltmeter measures a node whose output resistance is 500 kΩ. Estimate the loading error.
6. A network contains a dependent source. A student sets every source to zero before finding output resistance. What is wrong with the procedure?

DC-03 is now conceptually closed: topology determines which quantities are shared, Kirchhoff's laws constrain them, Ohm's law turns those constraints into series/parallel reductions and divider relations, and loading tells us when the simplified network must be rebuilt.