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What is the current through the 3 ohms, the current through the 4.5 ohms, and the volts of the second battery?

The ammeter in the figure reads 3.0A.

___3 ohms_____4.5 ohms__

| . . . . . . . . . | . . . . . . . . . . |

| . . . . . . . . 2 ohms . . . . . . |

+ . . . . . . . . .| . . . . . . . . . . +

9V . . . . . . . .| . . . . . . . . . .?V

bat . . . . . . .(A) . . . . . . . . .bat

- . . . . . . . . . | . . . . . . . . . . -

| . . . . . . . . . | . . . . . . . . . . |

| . . . . . . . . . | . . . . . . . . . . |

|___________|____________|

Update:

I tried working it and got the equivalent resistance in the left loop to be 5 ohms, and thus 1.8A going through the 3 ohms, but the computer says that is incorrect.

2 Answers

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Best answer

You will apply Kirchoff's rules to solve this problem.

Define the direction of flow in each of the two loops. We'll define the left loop as clockwise and the right loop as counterclockwise. I chose these directions because intuitively, those appear to be correct, but the fact is that you could define both as clockwise, too. It doesn't matter; if you choose the wrong direction, you'll just get a negative value for current.

Set up your equations for each loop:

E1 – i1r1 – i3r3 = 0. [Kirchoff's loop rule, eq.1]

E2 – i2r2 – i3r3 = 0. [Kirchoff's loop rule, eq.2]

i1 + i2 = i3. [Kirchoff's junction rule, eq.3]

Plug in values and solve the system for your unknowns:

9 – 3i1 – 6 = 0 [eq.1]

3i1 = 3

i1 = 1

1 + i2 = 3 [eq.3]

i2 = 2

E2 – 9 – 6 = 0 [eq.2]

E2 = 15

So you have 1.0A through the 3Ω resistor, 2.0A through the 4.5Ω resistor, and the second battery is 15V.

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Since this was answered 6 years ago and the image could not longer be generated, the key to using the above set of equations is using given value of the ammeter for R3 and 2 ohms resistor located on the wire in the middle as R3. Everything else falls into place after that.

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