Hey people,
I'm currently learning about electricity and becoming an electrical engineer.
I just don't understand why 62V flows through R1 and 18V through R2 and how they arrived at R2=7.75.
(1 votes)
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So either I look at something or the task is described incorrectly.
80 V are applied to the series connection of R1 and R2.
The series circuit R1 and the R2 flow 8 A
The heist R1 has 10 ohms and is passed through with 8 A.
U= I*R
U= 8A * 10 ohms
U=80 V
The complete 80 V is available at R1
R2 must have 0 ohms. If it were larger, no more 8A can flow in the series connection.
Where did you hate the job?
This is in the physics practice. I asked a few other people and they said the same thing. So I just ask the professor.
First of all: voltage does not flow, voltage is applied. Electricity flows.
You know that over R3 80V. Also above the series circuit R1 -R2. You know the total current, and there is how the current divides. This allows you to calculate the current through R3 as well as the current through R1 -R2.
For R3: from the known voltage (80V) and the current, you can calculate R3.
R2: From the given voltage and the current you can calculate the sum of R1 and R2. Then remove this value R1 and get R2.
There are several ways to solve this.
So you mean that?
R1+R2=80/8
10+R2 = 10
But then the 0
Sorry I hadn’t done physics for a long time and I’m trying to understand everything from my own.
I looked again at the task – something seems to be wrong. First of all, I would expect to be like you, R2 is 0Ω with me.
In the example, the one voltage of 62V over R1, I cannot understand. We have 8A and 10Ω, U=R*I, 80V.
Do it. If you find time, let me know.
Yes! I’m just asking my profesor.