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Solving a simple circuit

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Dankenstien10

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Assuming the switch is closed, solve the circuit to find the value of the amp meter and the value of Rx.

V1=30v(located at the bottom)
V2=10v

I'm unsure how to approach this because of the arrangement and how the two voltages interact. Is it treated like a series circuit? Do the voltages combine/split at the junctions.

My guess is split the voltages at the right and top junctions. Although it seems to easy to be correct:

Amp meter = 20/5 =4A
Rx = 20/4 = 5V

We have only just started electronics in class and have covered only the very beginnings on series and parallel circuits. So I'm sure its following the same lines, just need clarification on how to go about solving this. Any help appreciated!

Many thanks!
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The current is the same at all points in a simple series circuit.
The voltage across the five ohm resistor is V1 - V2, so 20V. 20V across 5R = 4A.

Rx has half the voltage of the 5R at the same current, so half the resistance; 2.5 Ohms.

Cross check: 4A * 2.5 Ohms = 10V
 
The current is the same at all points in a simple series circuit.
The voltage across the five ohm resistor is V1 - V2, so 20V. 20V across 5R = 4A.

Rx has half the voltage of the 5R at the same current, so half the resistance; 2.5 Ohms.

Cross check: 4A * 2.5 Ohms = 10V

Thanks for your reply, for some reason I was overthinking it as if the voltages were "batteries" and combined together in some way. Felt a bit silly upon realising actually they're just voltmeters ha. Using KVL I can see that V=V1-V2 =20v and the 10v across rx is given to get the resistance. Got there in the end!
 
If V2 was a battery then Rx cannot be calculated as the current supplied (or sunk) by the second battery is not known.

Mike.
 
If V2 was a battery then Rx cannot be calculated as the current supplied (or sunk) by the second battery is not known.

Mike.
Don't confuse things. V2 is a voltmeter. NOT A BATTERY.
 
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