# Circuit Analysis Question

Discussion in 'Homework Help' started by madhumt, Mar 6, 2014.

1. ### madhumt Thread Starter New Member

Mar 6, 2014
2
0
3A 125V circuit contains a 10W resistor.What resistance must be added in series for the circuit to have a current 0f 5A ?

2. ### Georacer Moderator

Nov 25, 2009
5,151
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Your post has been moved into its own thread, where it will draw more responses.

Please refrain from "hijacking" existing threads with tangent or off-topic questions. You can create new threads for new questions using the "New Thread" button on the upper left of the page.

3. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
What have you worked out thus far?

Do you know what happens when you add a resistor in parallel with another? What happens to the current? The voltage? Know any laws that might apply to resistors and the current through them and the village across them?

Please post what you have done.

4. ### BytetoEat New Member

Mar 5, 2014
25
5
3A 125V = 375W ..This circuit contains 10W resistor which is dissipating way over its power rating.

Furthermore, if the current is currently 3A, and you want to increase it to 5.. how in the hell would adding more resistance in series do that?

5. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
It doesn't say that the 125V source is applied across the resistor, just that that is the input to the circuit that the resistor is a part of.

6. ### BytetoEat New Member

Mar 5, 2014
25
5
Right, its as vague as can get. I'm going off the few details he said and guessing since he didn't tell us the topology..

If your whole circuit is drawing 3A at 125V, then use ohms law and the resistance is
125V/3A = 41.6 ohms.

Since you want to increase your current, you put a resistor in parallel not series. If you want to resist your current less, then you need more "paths" in your circuit loop. Putting a resistor in series would impede the electron flow(less current).

Since your current resistance is 41.6 ohms, if you added another 41.6 ohms in parallel you would double your current to 6A, because now your electrons have two paths of the same exact resistance essentially making it half as hard to flow through the circuit. (think of a constant water pressure having two hoses to flow through instead of just one and this will make more sense)

The formula for equivalent parallel resistances is
$\stackrel{R1xR2xR3x...Rn}{R1+R2+R3+...Rn}$

We know the end result has to draw 5Amps, so we can say 125/Req=5A

Then you can set this value Req=$\stackrel{41.6xR2}{41.6 + R2}$

7. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
You need to clarify things a bit. What is the circuit that you are starting with? When you say 3A 125V circuit, what do you mean. At first I thought you meant a 125V circuit that is capable of supplying up to 3A (just like I would say that I have several 15A 120V circuits in this house without making any claim that any one of them is actually presently supplying the full 15A). But that doesn't make sense if you are saying you want to increase the current draw above 3A.

As others have pointed out, adding any resistance in series will normally only reduce the current draw.

We really need more information -- as well as for you to show YOUR effort to solve YOUR homework problem -- before we can be of much help.