# dc series circuit problem

Discussion in 'Homework Help' started by pidge, Apr 26, 2010.

1. ### pidge Thread Starter New Member

Apr 26, 2010
1
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i have 3 resistors in a 329VDC circuit series R1=9V R2=25ohms R3=15ohms how can i find out the ohms value for R1 or what i'm thinking is it's a trick qu. and i need more values

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
Ohm's law is all you need.

3. ### TheWeasel New Member

Aug 28, 2009
4
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beenthere's right, there's no trick, you have all you need, and it can be worked out as follows;

The volts drop given across R1 is 9v which leaves the remaining 320v to be dropped across R2 + R3 (40 ohms).
I=V/R so 320/40 gives 8 Amps.

Knowing the current now permits calculation of the value of R1 needed to drop 9v at 8A ;
R= V/I or 9/8 which makes R1 = 1.125 ohm

As a double check in the complete circuit;
V/R = I or 329V/41.125R = 8A......so it all fits together.

Last edited: Apr 26, 2010
kingdano likes this.
4. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
To TheWeasel - it might be more instructive to get the OP to make some of the calculations and see some of the relationships himself. More learning goes on that way.

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5. ### kingdano Member

Apr 14, 2010
377
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i agree with beenthere - but for me it was always helpful to see an example worked through once in a while.

so kudos to both of you for the helpful posts to the EIT.