Ohm's Law Problem
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From: Rich Webb on 15 Feb 2010 06:35 On Mon, 15 Feb 2010 00:48:46 -0500, Spehro Pefhany <speffSNIP(a)interlogDOTyou.knowwhat> wrote: >On Sun, 14 Feb 2010 16:12:04 -0800 (PST), the renowned 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. >> >> V1 +8.4 >> | >> R1 = 240 >> | >> V2 .---------> 5 mA >> | >> R2 = 570 >> | >> V3 .---------> 5 mA >> | >> R3 = 100 >> | >> | >> GND > > > >(V1-V2)/R1 = 5mA + (V2-V3/R2 >(V2-V3)/R2 = 5mA + V3/R3 > >V1 is known (8.40V), so this is 2 equations in two unknowns and easily >solved for V2 = 5.169230769 V, V3=0.346153846 V. The Curse of the Calculator strikes again! I3 = 3.5 mA, V3 = 0.35 V, and V2 = 5.2 V. Hard to justify more than two significant figures here. -- Rich Webb Norfolk, VA
From: Spehro Pefhany on 15 Feb 2010 09:46 On Mon, 15 Feb 2010 06:35:19 -0500, the renowned Rich Webb <bbew.ar(a)mapson.nozirev.ten> wrote: >On Mon, 15 Feb 2010 00:48:46 -0500, Spehro Pefhany ><speffSNIP(a)interlogDOTyou.knowwhat> wrote: > >>On Sun, 14 Feb 2010 16:12:04 -0800 (PST), the renowned 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. >>> >>> V1 +8.4 >>> | >>> R1 = 240 >>> | >>> V2 .---------> 5 mA >>> | >>> R2 = 570 >>> | >>> V3 .---------> 5 mA >>> | >>> R3 = 100 >>> | >>> | >>> GND >> >> >> >>(V1-V2)/R1 = 5mA + (V2-V3/R2 >>(V2-V3)/R2 = 5mA + V3/R3 >> >>V1 is known (8.40V), so this is 2 equations in two unknowns and easily >>solved for V2 = 5.169230769 V, V3=0.346153846 V. > >The Curse of the Calculator strikes again! > >I3 = 3.5 mA, V3 = 0.35 V, and V2 = 5.2 V. Hard to justify more than two >significant figures here. Hey there, we wouldn't want some question of the correct value of the NINETH digit on the 'ol Agilent 3458A, now would we.. ;-) Best regards, Spehro Pefhany -- "it's the network..." "The Journey is the reward" speff(a)interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
From: life imitates life on 15 Feb 2010 10:22 On Mon, 15 Feb 2010 09:46:23 -0500, Spehro Pefhany <speffSNIP(a)interlogDOTyou.knowwhat> wrote: >On Mon, 15 Feb 2010 06:35:19 -0500, the renowned Rich Webb ><bbew.ar(a)mapson.nozirev.ten> wrote: > >>On Mon, 15 Feb 2010 00:48:46 -0500, Spehro Pefhany >><speffSNIP(a)interlogDOTyou.knowwhat> wrote: >> >>>On Sun, 14 Feb 2010 16:12:04 -0800 (PST), the renowned 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. >>>> >>>> V1 +8.4 >>>> | >>>> R1 = 240 >>>> | >>>> V2 .---------> 5 mA >>>> | >>>> R2 = 570 >>>> | >>>> V3 .---------> 5 mA >>>> | >>>> R3 = 100 >>>> | >>>> | >>>> GND >>> >>> >>> >>>(V1-V2)/R1 = 5mA + (V2-V3/R2 >>>(V2-V3)/R2 = 5mA + V3/R3 >>> >>>V1 is known (8.40V), so this is 2 equations in two unknowns and easily >>>solved for V2 = 5.169230769 V, V3=0.346153846 V. >> >>The Curse of the Calculator strikes again! >> >>I3 = 3.5 mA, V3 = 0.35 V, and V2 = 5.2 V. Hard to justify more than two >>significant figures here. > >Hey there, we wouldn't want some question of the correct value of the >NINETH digit on the 'ol Agilent 3458A, now would we.. ;-) > > > >Best regards, >Spehro Pefhany Nineth? Doth thee haveth a nineth digit?
From: Spehro Pefhany on 15 Feb 2010 10:49 On Mon, 15 Feb 2010 07:22:37 -0800, the renowned life imitates life <pasticcio(a)thebarattheendoftheuniverse.org> wrote: >On Mon, 15 Feb 2010 09:46:23 -0500, Spehro Pefhany ><speffSNIP(a)interlogDOTyou.knowwhat> wrote: > >>On Mon, 15 Feb 2010 06:35:19 -0500, the renowned Rich Webb >><bbew.ar(a)mapson.nozirev.ten> wrote: >> >>>On Mon, 15 Feb 2010 00:48:46 -0500, Spehro Pefhany >>><speffSNIP(a)interlogDOTyou.knowwhat> wrote: >>> >>>>On Sun, 14 Feb 2010 16:12:04 -0800 (PST), the renowned 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. >>>>> >>>>> V1 +8.4 >>>>> | >>>>> R1 = 240 >>>>> | >>>>> V2 .---------> 5 mA >>>>> | >>>>> R2 = 570 >>>>> | >>>>> V3 .---------> 5 mA >>>>> | >>>>> R3 = 100 >>>>> | >>>>> | >>>>> GND >>>> >>>> >>>> >>>>(V1-V2)/R1 = 5mA + (V2-V3/R2 >>>>(V2-V3)/R2 = 5mA + V3/R3 >>>> >>>>V1 is known (8.40V), so this is 2 equations in two unknowns and easily >>>>solved for V2 = 5.169230769 V, V3=0.346153846 V. >>> >>>The Curse of the Calculator strikes again! >>> >>>I3 = 3.5 mA, V3 = 0.35 V, and V2 = 5.2 V. Hard to justify more than two >>>significant figures here. >> >>Hey there, we wouldn't want some question of the correct value of the >>NINETH digit on the 'ol Agilent 3458A, now would we.. ;-) >> >> >> >>Best regards, >>Spehro Pefhany > > > Nineth? > > Doth thee haveth a nineth digit? Well, eight and a half, so not directly applicable in this particular case, but it's not totally ridiculous calculating DC voltages to 10 or 11 places when such tools are at hand. Certainly to 7 places. Best regards, Spehro Pefhany -- "it's the network..." "The Journey is the reward" speff(a)interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
From: Jim Thompson on 15 Feb 2010 11:11
On Sun, 14 Feb 2010 17:25:29 -0800, Jon Kirwan <jonk(a)infinitefactors.org> wrote: >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 Everyone is making it too difficult. Just write it down in sequence, the answer falls right out... http://analog-innovations.com/SED/OhmsLaw_SED_JustWriteItDown.pdf At MIT, I was spared (*) from Guillemin's obtuseness, I had Harry B. Lee for passive circuit analysis ;-) (*) I was awarded honors at the end of freshman year and went into course VI-B (just six of us, made for nice class sizes :-) ...Jim Thompson -- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food. |