Question

Simple circuit diagram problem; not really familiar with E&M. Posted below in a moment.

Answer 1

"What is the current through the 3 Ω resistor in the circuit of the figure (Figure 1) ? Hint: This is trivial."

http://session.masteringphysics.com/problemAsset/1256837/2/RW-25-28.jpg

Answer 2

(Crazily formatted question mark is supposed to be the Greek letter for Ohms.)

Answer 3

@ganeshie8 Maybe you understand this?

Answer 4

In my mind, the potential difference should be (mainly) due to the two emf's with positive terminals facing each other, and them "meeting" each other on the wire, the potential difference between the two of them just being 3 Volts. Apparently, the voltage across the resistor is 6 V, and I don't understand why that is.

A quick thing to think is that the 9V emf just entirely *DOES NOT AFFECT* the 3 Ohm resistor, but I can't find any reason to *actually* make that case, other than, "it's convenient".

Answer 5

Are you familiar with the Superposition principle? When there are two or more sources, solve the circuit considering one source at a time (short circuit rest of the voltage sources and open circuit rest of the current sources)...Ring any bells??

Answer 6

^that by itself doesn't really adequately explain why I should choose a particular source of current to ignore; if I happened to arbitrarily choose to ignore the 9V resistor, would I have gotten different results?

Answer 7

lol, mistakes everywhere. 6V EMF/battery, I mean.

Answer 8

you are both correct. If you apply the principle of superposition to the 6V battery then the 3 ohm resistor is shorted out and the 9V battery does not "see" it, i.e., it has no affect. When you apply the principle to the 9 V battery the 6V battery "see" both resistors but the 5 ohm resistor has no affect on the current in the 3 ohm resistor, right? When you connect a voltage source (of negligible internal resistance!!) the voltage across the resistance is defined and the current is fixed at V/R no matter what else is connected.

Please note that if a voltage source has significant internal resistance when you short out the voltage you must not eliminate its internal resistance for now instead of a dead short across the resistor you have a conductive path or shunt which causes the other voltage sources to "see" a resistance which produces an additional voltage drop which affects other components in the rest of the circuit..

Answer 9

It doesn't matter which emf source you choose first or how complex the circuit.

Answer 10

The 3 Ohm resistor has 6 volts across it. I = E/R = 6/3 = 2 Amps. Save superposition for more complicated circuits, this is Ohm's law.