ByFelbet
Hello,
I'm not making progress on a practice exam.
I've simplified the circuit, but the total resistance is a bit of a problem…
Can anyone help me with this?
Thank you.
(1 votes)
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(a)
(b)
Without internal resistance, the voltage between AB would be equal to the voltage of the source. However, a part of the voltage drops due to the internal resistance, since the entire current of the circuit flows through it. You did not give it, but you can apply the voltage divider because the internal resistance Ri is in series with the total resistance R_AB.
(c)
First you can determine the total current:
Now you’re dripping up the circuit you’ve summarized in a), something like that R_125 is still combined, but R4 and R3 are still separate.
Draw the circuit and you see that R4 and R125 are in series, i.e. through them the same current flows. Now you have two options.
variant 1: Power divider
variant 2: About the voltage
Since the current I4 is at the same time the current which flows through R125 and is the voltage across R4+R_125 =UAB, the following applies:
You’ve redrawn the circuit, you haven’t simplified it yet.
To do this, you combine step-by-step resistors that are in series or parallel.
So R1 and R2 (in row) to R1_2
Then R1_2 and R5 (parallel) to R1_2_5
Then take R4 (again in row), which then results in R1_2_4_5
Then the parallel connection with R3 and only a single resistor remains.
Thank you.
come to 62,96 ohms, so a) would be solved.
can you help me for b and c?
b) which voltage is between A and B at 6V and Ri of 10 ohms?
c) which current has the current flowing through the resistor R4?
With easylife2 switching, you can set up the following network coarse formulation:
Next, the parallel circuits are formulated, which leads to an unappetitious double break. However, this can be converted into a single fracture by a suitable extension. The rest is for the calculator.
“Useful” is here the currency with which is paid.