Question

An infinite ladder network of resistances is constructed with 1 ohm and 2 ohm resistors as shown in figure below. The 6 volt battery between A and B has negligible internal resistance.(i) Show that effective resistance between A and B is 2 ohms.
(ii) What is the current that passes through the 2 ohm resistance nearest to the battery?

Solution

The resistances used in the circuit are 1

Ω and 2 Ω

.
A supply voltage of 6 V is applied across A and B.
Let the given circuit be broken in parts as shown in fig. (a) and (b).

Since the circuit is infinitely long, its total resistance would remain unaffected even if we remove one mesh from it.
Let the effective resistance of the infinite network be R. The effective resistance of the remaining part of the circuit beyond CD is also R.
The circuit can be recombined as shown in fig. (b).

i) The resistance R

Ω

and 2

Ω

are in parallel.

Their combined resistance is

R' = \frac{2 R}{R + 2}

R’ is in series with remaining 1 Ω resistance.

Therefore, the total combined resistance is

\frac{2 R}{R + 2} + 1

which must be equal to the total resistance of the infinite network.

Therefore,

R = \frac{2 R}{R + 2} + 1 = \frac{3 R + 2}{R + 2}

or,

R^{2} + 2 R = 3 R + 2

or,

R^{2} - R - 2 = 0

∴ R = \frac{1 \pm \sqrt{1 + 8}}{2} = 2 Ω

We have just considered the positive value of resistance since, R cannot be negative.

ii) Now, applying Kirchhoff’s loop rule to the two neighbouring meshes in fig. (b), we get

1 x I + 2 I’ = 6 (

∵

R = 2Ω)

2 (I – I’) – 2 I’ = 0

From second equation we get,
I = 2 I’

∴ 4 I’ = 6

\Rightarrow

I’ = 1.5 A

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