Current Electricity 2 Question 27
27. An infinite ladder network of resistances is constructed with $1 \Omega$ and $2 \Omega$ resistances, as shown in figure.
The $6 V$ battery between $A$ and $B$ has negligible internal resistance.
(1987, 7M)
(a) Show that the effective resistance between $A$ and $B$ is $2 \Omega$.
(b) What is the current that passes through the $2 \Omega$ resistance nearest to the battery?
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Solution:
- (a) Let $R _{A B}=x$. Then, we can break one chain and connect a resistance of magnitude $x$ in place of it.
Thus, the circuit remains as shown in figure.
Now, $2 x$ and $x$ are in parallel. So, their combined resistance is $\frac{2 x}{2+x}$
$$ \text { or } \quad R _{A B}=1+\frac{2 x}{2+x} $$
But $R _{A B}$ is assumed as $x$. Therefore,
$$ x=1+\frac{2 x}{2+x} $$
Solving this equation, we get
$$ x=2 \Omega $$
Hence proved.
(b) Net resistance of circuit $R=1+\frac{2 \times 2}{2+2}=2 \Omega$
$\therefore$ Current through battery $i=\frac{6}{2}=3 A$
This current is equally distributed in $2 \Omega$ and $2 \Omega$ resistances. Therefore, the desired current is $\frac{i}{2}$ or $1.5 A$.
