Fill in the missing voltages and amps:

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- Apr 27th 2007, 07:26 PMDivideBy0Circuit
Fill in the missing voltages and amps:

- Apr 28th 2007, 07:29 AMtopsquark
The current in the main branch (that connected directly to the power supply) will all be the same, so we have that the two unknown currents in the main branch are still 12 A. (See "Tough Circuit2").

Now, the potential difference across an ammeter should (in theory) be 0 V. So the meter reading across the ammeter in the top right corner is 0 V. The same argument applies to the potential difference across the ammeter in the lower right corner. (See "Tough circuit3.")

We know the current in the bottom branch of the circuit (the one with only one 10 Ohm resistor) is 9 A. So what's the potential difference across it? V = IR, so V = (9A)(10 Ohm) = 90 V. (See "Tough circuit4.")

Now take a look at the junction on the left side where the 9 A runs into the 12 A. (The vertical line on the left side.) What is the unknown current running into this junction? Well, by tracing the direction of the currents we know that both the 9 A and 12 A branches have the current flowing upward here. So we have to ADD current to the 9 A to bring it up to the 12 A. Thus the unknown current coming into the junction is 3 A and is flowing to the left. (See "Tough circuit5.")

Finally, let's get the potential across the two 10 Ohm resistors. The two resistors are in series, so the equivalent resistance of the two is 10 Ohm + 10 Ohm = 20 Ohm. The current in this branch is 3 A, so the potential across the resistors is (3 A)(20 Ohm) = 60 V. (See "Tough circuit6.")

-Dan - Apr 28th 2007, 08:37 AMCaptainBlack
Now what bothers me about this is the potential accross the branch

with the single 10 ohm resistor is the same as that accross the branch with

the two 10 ohm resistors (as these branches are in parallel)

Or have I got confused somewhere?

According to my calculations the 9A should be 8A, that is the 12A should split into

8 and 4A

RonL - Apr 28th 2007, 09:01 AMtopsquark