Applying Kirchoff’s rules using voltages, not currents

Method:

• Assign an unknown voltage to every junction in the circuit. You can setthe voltage at one junction — whichever one you choose — to be anythingyou like, usually zero.

• For every pair of junctions which has a direct path between them (i.e.not passing through other junctions), obtain the current in terms of thevoltage difference between the two ends, using Ohm’s law etc. If any pathonly has a voltage source in it (no resistors), the voltage difference betweenthe two ends is fixed by the voltage source.

• Impose the condition that the total current flowing into each junctionis equal to zero. This gives a set of equations for the various junctionvoltages.

• Solve the equations to find the junction voltages.

• For each direct path between junction pairs, now that you know the voltagedifference between the ends, find the current.

Examples

Giancoli 26-41: Resistor networkTo find the resistance between the specified points, connect a voltage sourcewith voltage V0 between them and apply Kirchoff’s rules. Let us call the apexof the triangle d.

In the first case, there is no symmetry in the problem. If Va = V0 andVc = 0, the currents flowing into b from all three paths leading into it add up to(V0−Vb)/R+(0−Vb)/R+(Vd−Vb)/R = 0. The currents flowing into d from allthree paths leading into it add up to (V0−Vd)/R+(Vb−Vd)/R+(0−Vd)/R′ = 0.Combining these equations, we have

V0 + Vd − 3Vb = 0

V0 + Vb − 2Vd = VdR/R′. (1)

We do not apply Kirchoff’s junction rule to a or c because the current flowingalong the path between them with the voltage source (the path has zero resis-tance and zero voltage drop outside the voltage source, so that the current fromOhm’s law is 0/0). But the two equations are enough to obtain

Vd = 4V0R′/(5R′ + 3R)

Vb = V0(3R′ + R)/(5R′ + 3R). (2)

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Therefore the total current flowing into the triangle from a towards b, c and d,which is (V0 − Vb)/R + (V0 − 0)/R + (V0 − Vd)/R, is equal to

V0

R

[2R′ + 2R

5R′ + 3R+ 1 +

3R + R′

5R′ + 3R

]=

V0

R

8(R′ + R)

5R′ + 3R. (3)

This is, therefore, the current flowing into a from the battery. Therefore theeffective resistance of the triangle between a and c is R(5R′ + 3R)/[8(R + R′)].

You have to work with two unknown voltages. By contrast, if you had usedthe method in your book, you would start with 7 unknown currents (each pathin the triangle, and through the battery). You would apply Kirchoff’s junctionrules to reduce this to 4 unknown currents (one junction equation is alwayssuperfluous), and then use Kirchoff’s loop rule to obtain 4 equations for these4 currents. This is more complicated, but is certainly correct. You can do thisif you prefer.

In the second case, Va = V0 and Vb = 0. By symmetry, Vc = Vd. The totalcurrent flowing into d from a, b and c is

V0 − Vd

R+

0 − Vd

R+ Vd − VdR = 0 (4)

so that V0 − 2Vd = 0. By symmetry, applying Kirchoff’s rules to c gives thesame result. As before, we can’t apply the junction rule to a or b. From our oneequation we have Vd = Vc = V0/2, so that the total current flowing out from ainto the triangle is (V0 − V0/2)/R+ (V0 − V0/2)/R+ V0/R = 2V0/R. This mustbe the current flowing into a from the battery. Since connecting a battery ofvoltage V0 results in a current of 2V0/R, the resistance of the circuit is R/2.

Giancoli 26-42: Resistors and voltage sourcesThere are two junctions in this problem. Since all the three direct paths betweenthese junctions have resistors in them, not just voltage sources, we do not knowthe voltage difference between the junctions. Let the voltage at the junctionon the right be V1 and at the junction on the left be zero. The current flowingfrom right to left along the topmost path is V1/4Ω. The currrent flowing alongthe bottom path is (V1 − 3V )/0.45Ω, because of the voltage drop V1 from rightto left, 3 Volts occurs across the battery, so that the rest must be across theresistor. The currrent flowing along the middle path is (V1 − 2V )/0.45Ω. Allthese three have to add up to zero. This implies that V1, which is the voltagedrop across the resistor R, is 2.367 V. Current flows from right to left in theweaker battery, i.e. opposite to the direction it tries to push the current, becauseit is overridden by the stronger battery.

Giancoli 24-40: Capacitors and voltage sourcesWe avoided these problems before. The nice thing is that you can solve themapplying exactly the same method as above, except that you have to work withcharges instead of currents. If the two plates of a capacitor of capacitance Care at potentials Va and Vb, the charges on the two plates are Qa = (Va − Vb)Cand Qb = (Vb − Va)C, which are obviously equal and opposite.

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In the figure in your book, let the voltages at the four junctions be, fromtop to bottom, V, V1, V2 and zero. The charges on the capacitor plates thatare connected to the junction at voltage V1 are then (V1 − V )C2, (V1 − V2)C3

and V1C1. These must add up to zero. Similarly, the charges on the capacitorplates that are connected to the junction at voltage V2 are (V2 − V1)C3, V2C4

and (V2 − V )C5. Using C1 = C3 = C5 and C2 = C4 we have

(V1 − V )C2 + (2V1 − V2)C1 = 0

(2V2 − V1 − V )C1 + V2C2 = 0 (5)

from which V1/V = (C1+C2)/(3C1+C2) and V2/V = 2C1/(3C1+C2). Thereforethe total charge that has flowed out of the voltage source at terminal a, whichis on the capacitors C2 and C5, is (V − V1)C2 + (V − V2)C5 which is equal to

Q = V

[C2

2C1

3C1 + C2+ C1

C1 + C2

3C1 + C2

]= V

[C1(C1 + 3C2)

C2 + 3C1

]. (6)

The expression in square brackets is the equivalent capacitance.Notice that we applied a ‘junction rule’ (net zero charge) only at the junc-

tions that lead to capacitors on all sides. For the top and bottom junction,which are connected to terminals of the voltage source, we do not know howmuch charge has flowed in from the battery — in fact, that is what we aretrying to calculate — and so there is no junction rule. So even though there arefour junctions, there are only two junction rule equations. But this was enoughto find the two unknown voltages V1, V2. This is exactly what happened in theprevious examples about resistor networks.

Circuits consisting of capacitors and several voltage sources can be solved inthe same way.

Capacitors and resistors and voltage sourcesThings get complicated for these circuits. A simple case is when (as in yourhomework) all the capacitors are either uncharged or you wait until they arefully charged. The voltage drop across an uncharged capacitor is zero, andso it can be replaced with a wire (a short circuit). Although the two are notcompletely equivalent because current flows through the wire but not throughthe capacitor, since current flows all the way up to one plate of the capacitorand starts again from the other plate, as long as we are not interested in pokinginto the innards of the capacitor we can pretend that current is flowing throughit. The current flowing into a fully charged capacitor is zero, because otherwisethe charge on the plates would change with time, which it does not. Thereforeit can be replaced with a cut wire (an open circuit) through which no currentflows. Thus the circuit is reduced to a collection of resistors and voltage sources,but the reduced circuit is different initially from what it is after a long time.

To find the general solution as a function of time t, not just initially and aftera long time, we have to derive a set of differential equations for the circuit. Theequations are, in fact, easier to write down if you use loop currents instead ofjunction voltages, just like your book does. But to find a steady state solution,

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we use Fourier transforms, where one can work with junction voltages. (Thisdoes not work for the general time dependent solution.) In any case, this isbeyond the scope of this course.

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