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Circuit question

So there is series circuit. It first has 3 resistors connected in parralel and has one resistor on its own called R. And finally there two resistors connected in parralel with in the series circuit just like all the other resistors. The resistance value for all the resistors are the same. And the p.d. at terminal is 12v. So what would the p.d. be at R
Original post by Hiphop678
So there is series circuit. It first has 3 resistors connected in parralel and has one resistor on its own called R. And finally there two resistors connected in parralel with in the series circuit just like all the other resistors. The resistance value for all the resistors are the same. And the p.d. at terminal is 12v. So what would the p.d. be at R


Hi, not sure if this is correct as I am only in year 13 haha and not a teacher or anything.
If you get the solution could you let me know? So I know if what I'm doing is correct :smile:

So I used circuit rules that parallel components all have the same pd, and since the resistors all have resistance R then the current across them must be the same too
but since current splits up at branches by K1L, you can call the current across them 1/3 I, with I being the current supplied by the terminal
Then you have R on its own with a current of just I and a resistance of R
Then the two in parallel both have resistances R and currents of 1/2I, because again current splits up
So you can label above all the resistors their pds using V=IR, so since pd is same for the parallel resistors you can put 1/3IR for the first combo, then just IR for the one on its own, then 1/2IR for the other two.
So call the first combo V(3R), one on its own V(R) and last combo V(2R)
You can then set V(R) = 3 * V(3R) = 2 * V(2R), by the labels we made above the resistors earlier
Then you can use these to create an equation in terms of V(R) as K2L means they should all add to 12. You get
12 = V(R) + 1/3 V(R) + 1/2 V(R)
12 = 11/6 V(R)
So V(R) = 6.545454...

Hope this all makes sense? And let me know if you do get a proper solution :smile:

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