Ohm's Law Problem
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From: Bill Bowden on 14 Feb 2010 19:12 Ohm's Law Problem: Find the voltage at the 2 junctions of a 3 element voltage divider across a supply voltage of 8.4 volts. The two junctions of the divider both supply external current of 5mA. +8.4 | R1 = 240 | .---------> 5 mA | R2 = 570 | .---------> 5 mA | R3 = 100 | | GND A more basic problem can be solved using Thevenin's idea with only 2 resistors of say 570 and 240 ohms and a 6.9 volt supply, and 5mA of current from the single junction. +6.9 | R1 =240 | .----------> 5 mA | R2 =570 | GND The output impedance at the junction is the 2 resistors in parallel, or about 169 ohms. The open circuit voltage ignoring the 5mA is about 2.044 volts, and so the voltage drop on R1 (240) is 2.044 - (169 * .005) = 1.2 volts. But I don't see an easy way to apply Thevenin to the other case where there are 3 or more resistors and junctions with known currents from the junctions. Any ideas? -Bill
From: Fred Bartoli on 14 Feb 2010 19:23 Bill Bowden a �crit : > Ohm's Law Problem: > > Find the voltage at the 2 junctions of a 3 element voltage > divider across a supply voltage of 8.4 volts. The two > junctions of the divider both supply external current of 5mA. > > +8.4 > | > R1 = 240 > | > .---------> 5 mA > | > R2 = 570 > | > .---------> 5 mA > | > R3 = 100 > | > | > GND > > > A more basic problem can be solved using Thevenin's > idea with only 2 resistors of say 570 and 240 ohms > and a 6.9 volt supply, and 5mA of current from > the single junction. > > +6.9 > | > R1 =240 > | > .----------> 5 mA > | > R2 =570 > | > GND > > The output impedance at the junction is the > 2 resistors in parallel, or about 169 ohms. > The open circuit voltage ignoring the 5mA is > about 2.044 volts, and so the voltage drop > on R1 (240) is 2.044 - (169 * .005) = 1.2 volts. > > But I don't see an easy way to apply Thevenin to > the other case where there are 3 or more resistors > and junctions with known currents from the junctions. > > Any ideas? > Apply it independently for each 5mA sink, then use superposition. -- Thanks, Fred.
From: Tim Williams on 14 Feb 2010 19:26 Try superposition?? Thevenin is a two port, nondependent equivalent, so if you're trying to express something more multidimensional, it may not be the right approach. Could you explain a bit more about what you're trying to do? Tim -- Deep Friar: a very philosophical monk. Website: http://webpages.charter.net/dawill/tmoranwms "Bill Bowden" <wrongaddress(a)att.net> wrote in message news:5a6868d4-f680-4d3a-b6cf-97ee3a5e9c42(a)f15g2000yqe.googlegroups.com... > Ohm's Law Problem: > > Find the voltage at the 2 junctions of a 3 element voltage > divider across a supply voltage of 8.4 volts. The two > junctions of the divider both supply external current of 5mA. > > +8.4 > | > R1 = 240 > | > .---------> 5 mA > | > R2 = 570 > | > .---------> 5 mA > | > R3 = 100 > | > | > GND > > > A more basic problem can be solved using Thevenin's > idea with only 2 resistors of say 570 and 240 ohms > and a 6.9 volt supply, and 5mA of current from > the single junction. > > +6.9 > | > R1 =240 > | > .----------> 5 mA > | > R2 =570 > | > GND > > The output impedance at the junction is the > 2 resistors in parallel, or about 169 ohms. > The open circuit voltage ignoring the 5mA is > about 2.044 volts, and so the voltage drop > on R1 (240) is 2.044 - (169 * .005) = 1.2 volts. > > But I don't see an easy way to apply Thevenin to > the other case where there are 3 or more resistors > and junctions with known currents from the junctions. > > Any ideas? > > -Bill
From: Jon Kirwan on 14 Feb 2010 20:25 On Sun, 14 Feb 2010 16:12:04 -0800 (PST), Bill Bowden <wrongaddress(a)att.net> wrote: >Ohm's Law Problem: > > Find the voltage at the 2 junctions of a 3 element voltage > divider across a supply voltage of 8.4 volts. The two > junctions of the divider both supply external current of 5mA. > > Vc=+8.4 > | > R1 = 240 > | > Va a---------> 5 mA > | > R2 = 570 > | > Vb b---------> 5 mA > | > R3 = 100 > | > | > Vg GND ><snip> > Any ideas? Use nodal analysis. I've labeled a few things above. Vc=8.4, Vg=0, and you are interested in Va and Vb, which are the voltages at node a and node b, respectively. I assume the two 5mA currents are "outward" due to the arrows showing. It's easier to use these conductances: G1 = 1/R1 G2 = 1/R2 G3 = 1/R3 Now, examine node a. It's equation looks like: (1) Va*(G1+G2) + 5mA = Vb*G2 + Vc*G1 Before you say you may not be able to remember how to do this, or even understand how I got the above equation, just stop thinking about voltage _differences_ and think in terms of superposition (currents flow two ways, at once) instead. Current flows both inward and outward. The currents outward flow via the only three paths that exist: R1, R2, and the 5mA outgoing flow. The currents flowing inward flow via just two paths (we've already accounted for the outward current, by definition): R1 and R2. Equation 1 then must first start with this realization: "Current outward equals current inward." Which you must know, since it isn't possible for electrons to pile up in the node. Current in and current out must indeed match up. So what is left is to put the currents IN and the currents OUT onto the two sides. Equation 1's left side reads like this: "The current that spills outward from node a is simply Va times the conductance outward via G1 and G2 (since it flows outward via both, you add them), plus of course the 5mA that is also flowing outward." Think of Va, here, as any arbitrary voltage that mysteriously (without needing to know the surrounding voltages) _forces_ current outward via R1 and R2. Since the 5mA is _known_ by definition, Va doesn't interact with it. It just _is_. So just add that current since it is an outward going one and belongs on this half of the equation. Equation 1's right side reads like this: "The current that spills inward into node a is simply Vc times G1 plus Vb times G2." We've already accounted for the 5ma and it doesn't spill inward. So it doesn't appear on this side. However, the surrounding voltages _do_ force a reverse current via whatever conductances they have available to them. So, of course, you must include them on the right side. That explains the first equation you need. Now, examine node b. It's equation looks like: (2a) Vb*(G2+G3) + 5mA = Va*G2 + Vg*G3 However, since we know out of hand that Vg=0, we can drop that term: (2b) Vb*(G2+G3) + 5mA = Va*G2 Okay, now you have two equations and two unknowns. Solve them. From equation 2b we have: (3) Vb = (Va*G2 - 5mA) / (G2+G3) We can now substitute equation 3 into equation 1: (4a) Va*(G1+G2)+5mA=G2*(Va*G2-5mA)/(G2+G3)+Vc*G1 (4b) Va*(G1+G2)=Va*G2^2/(G2+G3)-G2*5mA/(G2+G3)+Vc*G1-5mA (4c) Va*(G1+G2)-Va*G2^2/(G2+G3)=Vc*G1-5mA-G2*5mA/(G2+G3) (4d) Va*(G1+G2-G2^2/(G2+G3))=Vc*G1-5mA*(1+G2/(G2+G3)) (4e) Va*(G1+G2-G2^2/(G2+G3))=Vc*G1-5mA*(1+G2/(G2+G3)) (4f) Va=(Vc*G1-5mA*(1+G2/(G2+G3)))/(G1+G2-G2^2/(G2+G3)) Having solved for Va, let's compute it. For the values you gave, I get slightly over Va=5.169V. Plugging that into equation 3, I get just over Vb=346mV. Jon
From: John Larkin on 14 Feb 2010 20:25
On Sun, 14 Feb 2010 16:12:04 -0800 (PST), Bill Bowden <wrongaddress(a)att.net> wrote: >Ohm's Law Problem: > > Find the voltage at the 2 junctions of a 3 element voltage > divider across a supply voltage of 8.4 volts. The two > junctions of the divider both supply external current of 5mA. > > +8.4 > | > R1 = 240 > | > .---------> 5 mA > | > R2 = 570 > | > .---------> 5 mA > | > R3 = 100 > | > | > GND > > Take the 8.4 volts, 240 ohms, and 5 mA and convert that into a voltage+impedance. That reduces to 7.2 volts behind 240 ohms. Add 570 and it looks like 810 ohms, 7.2 volts. Suck 5 mA out of that, and so on. Just work your way down. Or do what I increasingly do, sic Spice on it. John |