Two topics of this lesson, the infinite ladder and the cube of resistors, go beyond the Class 12 syllabus. They are standard JEE material, and they use only the ideas of the last lesson plus one new tool: symmetry.
A network with no series and no parallel
Four resistors form a diamond A C B D, and a fifth joins C to D. A cell is joined across A and B. No two of the five resistors are in series, because every junction has three wires. No two are in parallel either. This arrangement is the Wheatstone bridge, and it cannot be reduced by the last lesson's rules alone.
Suppose, though, that C and D happen to be at the same potential. Then the middle resistor has no potential difference across it and carries no current. A resistor with no current does nothing, so it can be removed. What remains is two series arms in parallel. The bridge is then balanced.
\[ I_{CD} = 0 \Rightarrow I_Q = I_1,\; I_S = I_2 \]
Why. with no current in the middle, the current \(I_1\) through \(P\) (A to C) goes on through \(Q\) (C to B). Likewise \(I_2\) through \(R\) goes on through \(S\).
\[ V_A - V_C = V_A - V_D \Rightarrow I_1P = I_2R \]
Why. C and D are at one potential, so the drops from A to each are equal.
\[ V_C - V_B = V_D - V_B \Rightarrow I_1Q = I_2S \]
Why. for the same reason, the drops from each to B are equal.
\[ \dfrac{P}{Q} = \dfrac{R}{S} \]
Read it. divide the equation of step 2 by that of step 3; the currents cancel.
Balance depends only on the ratios. Each arm divides the cell's voltage. C and D sit at the same fraction of the way down when the two arms divide it in the same ratio. The values of the resistors, the cell's voltage and the middle resistor do not enter.
In the sim, C stays near \(12\,\text{V}\) because the upper arm does not change. D rises as \(S\) grows. At \(S = 6\,\Omega\) both sit at exactly \(12\,\text{V}\), since \(2/4 = 3/6\). The needle centres and the galvanometer carries nothing. Below \(6\,\Omega\) the galvanometer's current runs from C to D; above it, from D to C.
Arms of \(2\,\Omega\) (A to C), \(4\,\Omega\) (C to B), \(3\,\Omega\) (A to D) and \(6\,\Omega\) (D to B), with \(5\,\Omega\) between C and D. Find the current in the \(5\,\Omega\) and the equivalent resistance between A and B.
Test the balance: \(P/Q = 2/4 = 1/2\) and \(R/S = 3/6 = 1/2\).
Balanced, so \(I_{5\,\Omega} = 0\). Remove it: \(2 + 4 = 6\,\Omega\) in parallel with \(3 + 6 = 9\,\Omega\).
\(R_{AB} = 6 \times 9/15\) \(= 3.6\,\Omega\).
The \(5\,\Omega\) carries no current, and the bridge is \(3.6\,\Omega\) between A and B.
Five equal resistors \(R\) in a bridge are balanced, since \(R/R = R/R\). Remove the middle one: two arms of \(2R\) in parallel give \(R\). The balanced bridge returns in the chapter Cells, Bridges and Measuring Instruments. There it measures an unknown resistance, as the meter bridge of the Class 12 practical.
A bridge has \(P = 10\,\Omega\), \(Q = 20\,\Omega\) and \(R = 15\,\Omega\). What value of \(S\) balances it, in ohms?
A bridge is balanced. The \(5\,\Omega\) resistor across its middle is replaced by a \(50\,\Omega\) one. What changes?
Rings of wire
A uniform wire bent into a ring has its resistance spread evenly round it. Between any two points of the ring, current can go either way round. The two arcs share both end points, so they are in parallel. Each arc's resistance is its share of the whole ring.
For a \(12\,\Omega\) ring, points a quarter of the way round split it into \(3\,\Omega\) and \(9\,\Omega\): \(3 \times 9/12 = 2.25\,\Omega\). Opposite ends of a diameter split it into two \(6\,\Omega\) halves, \(3\,\Omega\). In general a ring of \(R\) gives \(R/4\) across a diameter. For points that divide it \(1 : 2\), the arcs are \(R/3\) and \(2R/3\), giving \((R/3)(2R/3)/R\) \(= 2R/9\).
A square with a diagonal
Five \(6\,\Omega\) resistors form a square ABCD with the diagonal AC. Between A and C there are three separate paths: A B C of \(12\,\Omega\), A D C of \(12\,\Omega\), and the diagonal of \(6\,\Omega\). They are in parallel: \(1/12 + 1/12 + 1/6\) \(= 1/3\), so \(3\,\Omega\).
Between A and B the picture changes. The side AB, \(6\,\Omega\), is one path. The other runs from A to C and then down CB. From A to C, the diagonal \(6\,\Omega\) is in parallel with A D C, \(12\,\Omega\), giving \(4\,\Omega\). Adding CB gives \(10\,\Omega\). So \(R_{AB} = 6 \times 10/16\) \(= 3.75\,\Omega\).
A uniform ring of wire has a resistance of \(18\,\Omega\). What is the resistance between two points one third of the way round from each other, in ohms?
A ladder that contains itself
A ladder of resistors has \(1\,\Omega\) in each section of the upper rail and \(2\,\Omega\) in each rung, and it goes on for ever. Call its resistance between the input terminals \(x\). Cover the first section with your hand. What is left is again an endless ladder of the same sections, so its resistance is also \(x\). An infinite ladder contains a copy of itself.
So the whole ladder is the first \(1\,\Omega\) in series with the first rung, \(2\,\Omega\), in parallel with \(x\): \[ x = 1 + \dfrac{2x}{2 + x} \] Multiply by \(2 + x\): \(x^2 + 2x\) \(= 2 + x + 2x\), so \(x^2 - x - 2 = 0\). The roots are \(2\) and \(-1\). A resistance cannot be negative, so \(x = 2\,\Omega\).
With \(R\) in every rail section and every rung, \(x = R + Rx/(R + x)\). This gives \(x^2 - Rx - R^2 = 0\), whose positive root is \(x = (1 + \sqrt5)R/2 \approx 1.62R\). The golden ratio of geometry turns up in a heap of resistors.
An infinite ladder has \(3\,\Omega\) in each rail section and \(6\,\Omega\) in each rung. What is its input resistance, in ohms?
Symmetry does the algebra
Twelve equal resistors \(r\) sit on the edges of a cube. Find the resistance between a corner A and the far corner G. Kirchhoff's rules of the next two lessons would need many equations. Symmetry needs none. Seen from A, the three edges that leave A look exactly alike, and so do their far ends. So a current \(I\) entering at A splits equally, \(I/3\) in each.
Each current \(I/3\) reaches a corner such as B, where two further edges leave. They are alike too, so each takes \(I/6\). Near G the picture is the mirror image: three edges bring \(I/3\) each into G. Now walk from A to G along A B C G, through one edge of each kind: \[ V_A - V_G = \dfrac{I}{3}r + \dfrac{I}{6}r + \dfrac{I}{3}r = \dfrac{5}{6}Ir \] So the resistance across the body diagonal is \(5r/6\). Points that symmetry makes alike, such as B, D and E, are at one potential, like the middle points of a balanced bridge.
The same method gives the other two cases. Across a face diagonal the cube is \(3r/4\), and across one edge it is \(7r/12\). There symmetry is weaker, so the current split must be found with the junction rule of the next lesson as well.
Twelve resistors of \(12\,\Omega\) each form the edges of a cube. What is the resistance between two corners at the ends of a body diagonal, in ohms?
In 1833 the English mathematician Samuel Hunter Christie described a diamond of four wires for comparing resistances. Few people noticed. Ten years later Charles Wheatstone, a professor at King's College London, used it in his lecture on electrical measurement and gave Christie the credit. The instrument took Wheatstone's name anyway. Wheatstone is also remembered for the concertina and for an early electric telegraph.
- Removing the middle resistor of a bridge without checking the balance.Test \(P/Q = R/S\) first. An unbalanced bridge needs Kirchhoff's rules.
- Adding the two arcs of a ring as if they were in series.Both arcs join the same two points, so they are in parallel.
- Keeping the negative root of the ladder's quadratic.A resistance is positive. Keep the positive root.