# Finding the voltage of a simple circuit

I'm trying to understand the solution of a circuit wrote by my professor. The problem asks to find the voltage across A and B in the following circuit: The solution provided is the following (where || is the parallel of two resistors): $$V_{AB} = \frac{R_2 \mid \mid R_L}{R_1 + (R_2 \mid \mid R_L)}V_S = \frac{1}{1 + \frac{R_1}{R_L} + \frac{R_1}{R_2}} V_S$$ I understood that he calculated the equivalent resistor for the parallel of R_L and R_2, but I don't understand how the potential across R_2 could be the same as the potential across R_2 || R_L.

• It won't be the same. Apr 23 at 16:45
• @Andyaka, huh? Both $R_2$ and $R_L$ terminate at the same points, A and B, right? Apr 24 at 9:21
• R2 is not the same value as R2||RL <-- read the question Apr 24 at 9:23
• @Andyaka regardless of differing value, they share both nodes of the circuit and therefore have by definition the exact same voltage across them at all times.
– Mels
Apr 24 at 11:24

I don't understand how the potential across R_2 could be the same as the potential across R_2 || R_L.

Don't think of it as "the potential across R2". Think of it as "the potential between A and B". Since both R2 and RL are connected between A and B, the potential between A and B is across R2, and it is across RL, and it is across the combination of R2 and RL. It's also the potential across the combination of VS and R1 in series. Because those are all things connected between A and B.

In a direct parallel circuit, such as R2 and RL, from two elements to a zillion:

All of the voltage is across all of the elements all of the time.

In a series circuit, such as VS, R1, and the parallel combination of R2 and Rl, from two elements to a zillion:

100% of the current goes through 100% of the elements 100% of the time.

Note that the currents through R2 and RL are equal only if the two resistances are equal. But the sun of their currents always equals the current through R1.

This is a very common way of measuring or monitoring the current through something - an LED, a motor, whatever. Place a small resistor in series with it, small enough not to change the circuit significantly, and measure or scope the voltage across the resistor.

To expand on Andy's comment -

The equation given as the solution is valid only for the schematic as shown. It is not correct if any part of the circuit changes. For example, disconnecting RL so it no longer is in parallel with R2 is a significant change. The overall loop resistance will be less, so the loop current will decrease. More importantly, ratio of R1 to 'everything-to-the-right-of-R1' will change. R2 by itself presents a higher resistance to R1 than R2||RL, so Vab will increase.

Run the numbers. Assign arbitrary values to all elements and use the parallel resistor equation and Ohm's Law to calculate things. For example, R1=3, R2=4. RL=12, Vs=6.

A lot of the answers here are correct, and I would just like to elaborate from the perspective of very clearly remembering when I was beginning and I got confused by this very thing. The voltage across R2 is the same as the voltage across RL. The trick is realizing that every place along the wiring lines connecting the rightward end of R1 to the top ends of R2 and RL is exactly the same. The same is true for the wiring lines connecting the bottom ends of R2 and RL to the negative terminal of Vs. The location of points A and B along those wiring lines is therefore arbitrary, and their placement "in line" with R2 is meaningless.

This is the very foundation of the computation of R2||RL in order to "solve" the circuit. If you have access to some basic components and a multimeter, you can test this yourself. In practice, real wires have a small resistance per unit length, and so long wires can increase the resistance on a circuit branch and change the voltage. But, this circuit diagram is a theoretical representation, and those wiring lines have no resistance of their own. No resistance = voltage stays the same everywhere along those lines. I hope this is helpful! Circuits 1 was one of my toughest courses, but also the one that convinced me to be an EE once it finally clicked.

According to second Kirchoff rule, you have two loops, the first from source and across R1 and R2 and the second across R2 and RL.

VS - R1 × I1 - R2 × I2 = 0
R2 × I2 - RL × IL = 0 (this second eqution is the quick answer)

Also apply the first Kirchoff rule to node A.

I1 = I2 + IL

You will shortly come up with a solution for your question using simple algebra.

Since you haven't declared any resistance or inductance between R2 and RL, the connection is assumed to be ideal and any voltage difference will be negligable.