Is the ammeter reading 2A and the power absorption 100W or did I calculate this incorrectly?
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3\$\begingroup\$ Your calculation is incorrect (well your answer is incorrect). I don't see anything that constitutes a calculation. \$\endgroup\$– Andy akaCommented Sep 12, 2020 at 16:09
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\$\begingroup\$ ... but it's hard to help you because you don't show us how you came up with the number. \$\endgroup\$– Marcus MüllerCommented Sep 12, 2020 at 16:09
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\$\begingroup\$ You have assumed that all of the 2 A is going through the resistor. If that is the case what is the voltage across it? If that voltage is on one side of the ammeter and 12 V is on the other what happens? \$\endgroup\$– TransistorCommented Sep 12, 2020 at 16:09
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1\$\begingroup\$ Should I calculate the power as 25ohm and 12V to get a power loss of 5.76W? Since the Current would be flowing through the resistor. The ammeter will then read 2-0.48A which is 1.52A. \$\endgroup\$– King of ZimbabweCommented Sep 12, 2020 at 16:18
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2 Answers
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Should I calculate the power as 25ohm and 12V to get a power loss of 5.76W? Since the Current would be flowing through the resistor. The ammeter will then read 2-0.48A which is 1.52A.
Because the ammeter's internal impedance can be assumed to be zero, there has to be 12 volts across the 25 ohm resistor. That means the power in the 25 ohm resistor is: -
$$\dfrac{12^2}{25} = 5.76\hspace{0.2cm}\text{watts}$$
And the current into the resistor is \$\dfrac{12}{25}\$ amps = 0.48 amps.
The rest of the current from the 2 amp source flows into the 12 volt source via the ammeter i.e. 1.52 amps.
answered Sep 12, 2020 at 16:34
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\$\begingroup\$ Thanks. That was the answer I ended up with too after relooking at it. \$\endgroup\$ Commented Sep 13, 2020 at 6:56
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If you use the water through pipes analogy, you will easily see that you are missing something. And the task is even telling you to consider the parallel "pipe" with that 25 ohm resistor. Work from first principles, intuition will guide you through this, and be prepared to consider the concept of "conductivity" as the inverse of "resistance". In parallel paths, their conductivity is additive. In serial paths, their resistance is additive. This all follows just by intuition (and if you don't have that intuition yet, then develop it, thinking of water through garden hose or pipes).
UPDATE: I was myself confused about the 2A symbol. If that's a current source then I was going the wrong way. I thought it was just an indication that there is 2A flowing though that branch driven by the 12 V source. Anyway, current through parallel paths is still additive.
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\$\begingroup\$ Should I calculate the power as 25ohm and 12V to get a power loss of 5.76W? Since the Current would be flowing through the resistor. The ammeter will then read 2-0.48A which is 1.52A \$\endgroup\$ Commented Sep 12, 2020 at 16:18
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\$\begingroup\$ Do you think current through parallel paths is additive or not? Think of water through hoses, if one hose transports 2 gallons per minute, and you have another parallel hose, both draining in a bucket of 2 gallons, do you think that bucket fills slower than in 2 minutes or faster? If one hose fills it in 2 min, why would two hoses fill it slower? \$\endgroup\$ Commented Sep 12, 2020 at 16:22
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\$\begingroup\$ One hose would fill it slower due to a restriction (the resistance). This is my new calculation: imgur.com/a/c09OuKu I figured that power will be absorbed so current will be going through the resistor. I calculated power by using ohm's law with the resistance and voltage. Voltage would be the same as it is in parallel. \$\endgroup\$ Commented Sep 12, 2020 at 16:25
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\$\begingroup\$ I just noticed that the branch ammeter showing 2A with an arrow pointed to the positive side. I find that very confusing. I wonder if that tricked you into a wrong way. The arrow should go from + down to - or ideally not show an arrow. The power through the 25 ohm resistor is something I would calculate only at the very end as P = U * I. You want to know how much current goes through the 25 ohm and then add those currents, because you still seem to subtract them. \$\endgroup\$ Commented Sep 12, 2020 at 16:40
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\$\begingroup\$ I misunderstood the 2A symbol. If that's a current source going against the 12 V source, then I suppose you've been going in the right direction. \$\endgroup\$ Commented Sep 12, 2020 at 16:46