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?
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?
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.
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".
simulate this circuit – Schematic created using CircuitLab
The first measurement the meter will read 3.33V.
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.