Calculate the voltmeter reading for this circuit with 2 sources and 2 resistors

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The discussion revolves around calculating the voltmeter reading in a circuit with two sources and two resistors. Initial calculations suggest a reading of 8 V based on a current of 2 A, but a participant argues the correct reading should be 12 V. The calculations involve determining the currents returning to each cell and the total current entering the voltmeter. Ultimately, the consensus leans towards 8 V being the correct reading despite the differing opinions. The thread emphasizes the importance of the original poster doing the majority of the problem-solving work.
songoku
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Homework Statement
Please see below
Relevant Equations
V = I.R
1685171477692.png


I find the current first, which is 2 A. Then the voltmeter reading would be 10 V - 1 x 2 = 8 V.

But the answer is 12 V.
I don't understand why it should be 10 + 1 x 2 = 12 V

Thanks
 
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I agree with your answer. It obviously cannot be more than 10V.
 
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Thank you very much haruspex
 
Yes, I agree, 8V is correct.
 
Current returning to cell A = x
Current returning to cell B = y
Current entering the voltmeter = z

10- 1*x=0
4- 2*y = 0
z = y + x

X=10
Y=2
Z=12

U=RI

1/R=1/1+1/2
1/R=1,5
R=2/3

U = (2/3)*12 = 8

The reading is 8 Volts.
 
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Welcome to PF, @erpelkon

erpelkon said:
Current returning to cell A = x
Current returning to cell B = y
Current entering the voltmeter = z

10- 1*x=0
4- 2*y = 0
z = y + x

X=10
Y=2
Z=12

U=RI

1/R=1/1+1/2
1/R=1,5
R=2/3

U = (2/3)*12 = 8

The reading is 8 Volts.
It is okay to offer alternate solutions to old schoolwork threads where the OP has already solved the problem. Please just keep in mind that in active threads, the OP must do the bulk of the work. We cannot offer solutions to them until after they have solved their schoolwork problem. Thanks. :smile:
 
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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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