# Simple Circuit Diagram (Ohm's Law Application)

by exitwound
Tags: application, circuit, diagram, simple
 P: 292 1. The problem statement, all variables and given/known data Consider the circuit shown below. (a) Determine the current reading of the ammeter. (b) Determine the voltage reading of the voltmeter. (c) Determine the power dissipated by the 500 W resistor. What would be the first steps in attempting this? There are differing resistances on the three branches of the circuit. Therefore, I have different amounts of current on each branch. As a result, I can't figure out the voltage drop across the first resistor. I don't know how to proceed.
 HW Helper P: 6,207 You can use Kirchoff's laws OR find the combined resistance, then find the total current and go from there.
 P: 292 I don't know how to find the total resistance in this situation. I know Resistance in series = R1 + R2 + R3 + ... Rn resitance in parallel is 1/R = 1/R1 + 1/R2 + 1/R3 + ... 1/Rn I don't know where to do the equivalent resistance substitutions in this diagram.
 P: 128 Simple Circuit Diagram (Ohm's Law Application) A parrallel circuit is like series circuits joined together. How many series loops do have in that circuit? What is the total p.d in each branch of this parallel circuit. What is the current going thorough the 330ohm resistor? What is the current going through the 250 ohm resistor (be careful here)? You don't need to combine resistors really for this question. What equation must you use to find the above? The the final bit is easy.
 P: 292 Ohm's Law. V=IR. I don't know how to start the problem. Having a voltmeter on a branch with no resistance is bugging me. Having an ampmeter measuring the current on one branch is something else I can't figure out. You say I don't have to combine resistances, but how can I figure out the current if I don't?
Mentor
P: 40,658
 Quote by exitwound Ohm's Law. V=IR. I don't know how to start the problem. Having a voltmeter on a branch with no resistance is bugging me. Having an ampmeter measuring the current on one branch is something else I can't figure out. You say I don't have to combine resistances, but how can I figure out the current if I don't?
A voltmeter is a very high input impedance measuring device. So you can treat the voltmeter as an (open or short?) circuit....

A current meter is a very low input impedance measuring device, so you treat it as a ? circuit...

Once you make those substitutions, you can combine the parallel resistances on the right side, figure out the total current, then the voltage across the parallel resistances, then the current through each leg. Give all that a try, and post your work here.
P: 292
Well, you can combine the resistances on the right side to 142.24 Ohms, then add it to the serial connection with the 500 Ohms for a total of 642.24 Ohms of resistance in the circuit.

Using that, we can find that the current flowing through the circuit as a whole is 9.34e-3 Amps.

Then we find the voltage drop across the first resistor. (9.34e-3)(500)=4.67V.

Use the 1.33 Voltage on the 250Ohm resistor I=V/R=532mA.

The analogy that seemed to help me the most was from an online friend. And it's funny, I'll probably try to put all my faith into Vikings from now on:

 (5:02:29 PM) : well, here's my wicked viking analogy. (5:03:06 PM) : think of the electrons as a horde of screaming vikings descending on a village of poor, unfortunate, yet sexy, irish womenfolk with attractive brown eyes. (5:03:27 PM) : now, these women, being irish, are cunning (5:03:41 PM) : so they cleverly erect a series of barries to protect them from the pillaging hordes. (5:04:06 PM) : now, the first time the vikings attacked, they put up three in a row (5:04:09 PM) : each one higher than the last (5:04:16 PM) : but the charging vikings were having none of it (5:04:22 PM) : and they burst on through each wall. (5:04:37 PM) : but they got more and more tired (5:04:57 PM) : so the first wall was pretty short and lame, it was just some bales of hay (5:05:17 PM) : they got through it and only lost 1.5 V of momentum (5:05:24 PM) : but getting rid of it wasn't much work. (5:05:38 PM) : the second wall was a lot better, it took them twice as much energy to get through it (5:05:44 PM) : so they lost 3 V of momentum (5:05:57 PM) : but they had to move a bunch of boulders out of their way - it was hard work! lots of work done here. (5:06:11 PM) : the last wall was made of SOLID AWESOME (5:06:33 PM) : but the vikings were undaunted and broke on through anyway; those irish girls were extremely alluring (5:06:48 PM) : but it slowed them down the rest of the way (5:06:54 PM) : and they had to do a crap ton of work to get rid of the wall (5:07:09 PM) : so they were too zoned out for rape & pillage anyway by the time they got there (see, clever irish chicks) (5:07:18 PM) : so they just limped back to their longboats.
 Mentor P: 40,658 How in the world is that quote relevant here?
 P: 292 It's how he described current, voltage, and resistance. He doesn't subscribe to the water analogy either because he said it's more problematic than helpful. I can't use the water analogy either. It just doesn't make sense in my head. But when he described it as Vikings, it was a little easier to understand what he meant. Why? I have NO idea. but it worked for now.
 P: 52 Am I wrong to say that the voltage stays constant throughout the entire circuit?
 Sci Advisor P: 4,016 Am I wrong to say that the voltage stays constant throughout the entire circuit? Yes, that would be wrong. Maybe the meters are freaking you out. You have a resistor in series with two resistors in parallel. You know how to work out the equivalent resistance of two resistors in parallel. So then you have two resistors in series. Can you work out the voltages across them? So, what would the voltmeter read? All of the resistors now have a known voltage across them. What current would flow through the 250 ohm one? What would the ammeter read? What power would be dissipated in the 500 ohm resistor?
 P: 52 I know very little about circuits, but isn't there something that stays constant throughout the entire thing? Since you say it's not the voltage, could it possibly be the current?
 Sci Advisor P: 4,016 Paulo.... Yes, in a series circuit, the current in all components is the same. In this case, however, there were parallel resistors that were not equal, so the currents in these were not the same. This is an excellent problem if you would like to have a go at solving it. Why not try if you have an interest in circuits? Send me personal mail if you have need help with it.
P: 1
 Quote by exitwound 1. The problem statement, all variables and given/known data Consider the circuit shown below. (a) Determine the current reading of the ammeter. (b) Determine the voltage reading of the voltmeter. (c) Determine the power dissipated by the 500 W resistor. What would be the first steps in attempting this? There are differing resistances on the three branches of the circuit. Therefore, I have different amounts of current on each branch. As a result, I can't figure out the voltage drop across the first resistor. I don't know how to proceed.

is it 4.5 volts!! across 500.. 250 and 330 is 1.5... current in 500 is .009 - - 250 is .006 and 330 is .004..

series

Et=E1+E2+E3.....
It=I1=I2=I3.......

Parallel

Et=E1=E2=E3......
It=I1+I2+I3.......

just a guess.... Use kirchoff's law!!
 Sci Advisor P: 4,016 You would work out the combined resistance of the 330 ohm and 250 ohm resistors. These are in parallel. Call this R. (but work it out exactly) Now add this resistor instead of the 330 and 250 ohms. So, you get 500 + R. Now you can work out the total current. It is = E / R or (6 volts / 500 + R) Then you can work out the voltage across R. This is the answer to part B. This is the same voltage as across the 250 ohm resistor. So, you can work out the current in this resistor. This is the answer to part A. This thread is nearly a year old and the answers were actually given in earlier posts, as well as explanations of how to solve it.

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