2
\$\begingroup\$

enter image description here

So, two resistors are in series. So \$ i = \frac{10V}{3k} = 3.33mA \$

So Voltage across R1 should be: \$ V = i \cdot R_1 \$

So it should be 3.33V, no?

Why is it 10V?

\$\endgroup\$
2
  • 2
    \$\begingroup\$ You are confusing potential with potential difference. \$\endgroup\$
    – Bart
    Oct 19, 2017 at 6:42
  • \$\begingroup\$ Voltage is always measured between two points, and if only one point is specified, it can be automatically assumed that the other point is ground. In this case, from that point in the picture to ground is exactly same as the source voltage. Now if you put point a and b at terminals of the resistor - then you get 3.33 V. \$\endgroup\$ Oct 19, 2017 at 8:44

3 Answers 3

4
\$\begingroup\$

The voltage across R1 is the difference between the voltage at the top of the resistor and the voltage at the bottom or the resistor. The voltage at the top is fixed to 10V because it is directly connected to the power supply (there is nothing in between for a voltage drop to develop over).

The voltage between R1 and R2 is 10V - 3.3V or 6.7V. If you calculate the voltage drop across R2 you will get a result of 6.7V confirming this result (the bottom is at 0V because it is the same node as the ground node of the power supply).

When you measure single node voltages in a simulation like this, they are always relative to ground.

\$\endgroup\$
2
  • \$\begingroup\$ @DeanFranks is it coincidence that the node between R1 and R2 is 6.7V just like voltage across R2 is 6.7V, it's not always gonna be the case, is it? Like what, voltage between R1 and R2 is whatever across R2??? \$\endgroup\$
    – Jack
    Oct 19, 2017 at 3:31
  • \$\begingroup\$ Yes, the bottom leg of R2 must be at 0V (it is connected to the ground node). The potential difference across R2 (3.33mA, 2k) is 6.67V, so the top leg must be 6.67V relative to ground. As Trevor described in his schematic, the voltage drop across a resistor is just the difference between the two terminals on the resistor. You have to use other methods to determine the voltage or one or the other legs relative to ground in order to find all the voltages. \$\endgroup\$ Oct 19, 2017 at 4:20
4
\$\begingroup\$

You are getting confused by voltages and where things are measured from.

The voltage at the top of R1 is 10V relative to the indicated ground. The voltage across R1 is 3.33V.

Maybe that still confuses you.

Think about using a voltmeter. If I tell you to attach it across R1 you will put one lead at the top of R1 and the other at the bottom on R1. Note that would be a different measurement from if I asked you to measure the voltage at the top of R1. The latter would imply.."from ground".

schematic

simulate this circuit – Schematic created using CircuitLab

The first measurement the meter will read 3.33V.

\$\endgroup\$
3
  • \$\begingroup\$ strange, in circuitlab, if I put mouse cursor on the node between R1 and R2, it'll show 6.67V, why is that? It looks like 6.67V is across R2. \$\endgroup\$
    – Jack
    Oct 19, 2017 at 3:28
  • 1
    \$\begingroup\$ @Jack Because the measurement is referenced to ground. So, if you put the cursor between R1 and R2, you are reading the voltage across R2. If you put the cursor at the top of R1, you are reading the voltage across R1 and R2. \$\endgroup\$ Oct 19, 2017 at 3:37
  • \$\begingroup\$ @Jack and notice that your measurements agree with Trevor’s schematic. If there is 10 V total, and 3.33 V across R1, that leaves 6.67 V across R2. \$\endgroup\$ Oct 19, 2017 at 3:40
1
\$\begingroup\$

Voltage is always a quantity between two nodes. For example, when you use your multimeter, you must use both wires to measure a voltage. You cannot simply use one and let the other hang.

If some one asks you, what is the voltage in a certain node of a device, he/she (usually) means what is the voltage compared to ground. Just remember, voltage is always a difference between two potentials.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.