Question

Q.) An infinite ladder network of resistance is constructed with 19 and 20 resistances, as
shown in figure.

What is the current that passes through the 2 ohm resistor nearest to the battery?​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

1.5A

It's a long one but try to understand it. Hope it helps.

Explanation:

Let the equivalent Resistance of the infinite network ladder be 'x'.

In this infinite ladder, consider the branch of 1 Ω and 2 Ω which is nearer to the 6 volts.

Except for that branch, the resistance of the remaining branches is 'x'.

[See if we take a glass of water from a heavy ocean, it makes no change. Similarly, if we consider a branch from an infinite network of resistance, it makes no change in the resistance of the network.]

Hope you understand here. The below method helps to find the equivalent resistance of the entire infinite network connections.

So

The connection will be like 2 Ω in the first branch is parallel to the x ohm of the infinite circuit.

⇒R(eff) = 2x/(2 + x)   [∵ If two resistors x and y are in parallel, then

                                     1/R(eff) = 1/x + 1/y  ⇒ R(eff) = xy / (x + y)]

R(effective) and 1 Ω  are in series.

⇒ R(eqv) = R(eff) + 1 Ω

⇒ x = 2*x/(2+x) + 1        {∵R(eqv) =x Ω  for this infinite ladder}

⇒x = (2x + 2 + x)/(2 + x)

⇒x*(2+x) = 3x + 2

⇒x² + 2x = 3x + 2

⇒x² + 2x - 3x - 2 = 0

⇒x² - x - 2 = 0

⇒x² - 2x + x - 2 = 0   [By factorization method]

⇒x(x - 2) + 1(x - 2) = 0

⇒(x - 2)(x + 1) = 0

∴x = 2 or x = -1

Resistances cannot be negative so x ≠ -1

∴x = 2 Ω

Effective or equivalent resistance of this infinte network is 2 Ω.

R = 2 Ω , V = 6V

From Ohm's law

V = iR

⇒6 = i*2

⇒i = 6/2

∴i = 3A

3 amp current flows through this infinite ladder.

In series, the current is constant.

So, 3 amp current flows in 1 Ω resistor and the parallel connection of 2 Ω  and 2 Ω. (refer (1) as x = 2 Ω )

Potential drop at 1Ω  resistor.

From Ohm's law,

V = iR

⇒V = 3*1 = 3 volts                       [∵ i = 3A and R = 1Ω]

Out of 6 volts, 3 volts are used by 1 Ω  resistance.

So the remaining 3 volts are used by the parallel connection.

As equal resistances are connected in parallel ( 2Ω  and 2Ω ), the current (i = 3A) will be split equally into the two branches.

Current in the two branches = i/2 = 3/2 = 1.5 A

∴The current in the 2Ω  resistance nearest to the battery = 1.5 A