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- Thread starter sai2020
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alphysicist

Homework Helper

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When you write out Kirchoff's loop equation for a circuit with only resistors and (ideal) batteries, if you know the values of the battery voltages and the resistors, the only unknowns in your equations are the currents. And the currents in the loop equations only show up when your loop passes through a resistor.

So, looking at your picture, can you start at the resistor, and find some path, following wires through the circuit, that only passes through batteries and ends up back at the resistor? If you can, then writing the loop equation for that path would only have one unknown--the current in the resistor that you are looking for. So what do you get?

- #3

shengzhoumi

- 2

- 2

This current will always pick the easiest way to flow, so you know that the loop will have no resistors. If you look at the diagram, there is only one path with no resistors except for resistor R.

When you finish drawing the loop, the loop should look like a screwed up j. It should have 4 batteries too.

After that, just use V=iR to calculate the current.

Happy Solving. :)

- #4

sushrutphy

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Hi...sushrut here...

Is the answer coming out to be 2 coulombs?

Is the answer coming out to be 2 coulombs?

- #5

shengzhoumi

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ya, ur right...good job

- #6

sharkasm

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- #7

supratim1

Gold Member

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look properly, there is a loop with only that resistor and batteries.

- #8

- #9

mkrishnan

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but won't the presence of other batteries and resistors affect the current flow in the resistance 'R'? for example when we use Maxwell's loop current rule; if current through one loop is i1,and in another loop be i2,then current through a common resistor between the loops is given by (i1-i2)!

wont stuff like this happen in this circuit? if not why?!

wont stuff like this happen in this circuit? if not why?!

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- #10

Whakataku

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To those of you who don't see from the implication, I shall expand.

Look carefully at the circuit, the three emf on the left of the circuit are moving the charge in the clockwise direction, and the one emf on the right is the opposite direction.

So for the sum of emf, don't forget to make the opposing emf have the opposite sign.

Then i = (*ε *net)/R is self-explanatory.

The devil is in the details :)

Look carefully at the circuit, the three emf on the left of the circuit are moving the charge in the clockwise direction, and the one emf on the right is the opposite direction.

So for the sum of emf, don't forget to make the opposing emf have the opposite sign.

Then i = (

The devil is in the details :)

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- #11

agnivesh

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but how is it that when 4 batteries of 4 v each r connected...u get the net emf to be 8 v

- #12

agnivesh

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ooops sorry...now i understood...that battery connected in opposite order will hav opposite sign

- #13

mdm1984

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- #14

gneill

Mentor

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It doesn't matter what's in parallel with an ideal source (unless it's another ideal source with a different potential, and then FZZZSST! BANG!). Nothing in parallel with that resistor will change the potential across it that's set by the indicated ladder of ideal voltage sources.

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