TIFR CSE 2012 | Part B | Question: 16

Question

Consider a complete binary tree of height $n$, where each edge is one Ohm resistor. Suppose all the leaves of the tree are tied together. Approximately how much is the effective resistance from the root to this bunch of leaves for very large $n$?

• A. Exponential in $n$.
• B. Cubic in $n$.
• C. Linear in $n$.
• D. Logarithmic in $n$.
• E. Of the order square root of $n$.

Answer 1

Sum of resistors when in series $r_{serial} = r_{1}+r_{2}$

Sum of resistors when in parallel, $\dfrac{1}{r_{parallel}}= \dfrac{1}{r1} + \dfrac{1}{r2}$
                                             
$\implies r_{parallel} = \dfrac{r_{1}\cdot r_{2}}{r_{1}+r_{2}}$

Every node siblings are in parallel and the sum of each level are in series and all node of the last level are tied so all are in series.

So, total sum of resistor $=$ root $ + $ total of level $2 \ + $ total of level $3 + \ldots + $ total of level $n-1 +$ total of level $n$ (in series)
$\begin{array}{cc}\textbf{ Level}&\textbf{Number of nodes}\\ \hline
1&1 (2^0)\\
2&2 (2^1)\\
3      &              4  (2^2)\\4                     &            8 (2^3)\\.\\.\\n-1                    &          2^{n-2}\\n                         &        2^{n-1}\\
\end{array}$

So, total sum of resistor $=$ root $+$ total of level $2 \ +$ total of level $3 + \dots.+$ total of level $n-1 + \ $ total of level $n$ (series)
$\qquad= 1 + \frac{r}{2} + \frac{r}{4} + \frac{r}{8} + \frac{r}{16} + \dots + \frac{r}{2^{n-2}} + (\underbrace{r + r + r + r + \dots +r}_{2^{n-1}  \ times}) $ (here $r = 1)$
$\qquad = 1 + \left\{ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{n-2}}\right\}(\text{decreasing GP}) + {(\underbrace{1+1+1+\ldots+1}_{ 2^{n-1}  \ times})} $

$\qquad \approx (2^n)  \ \text{approx}$

Option $(A)$ is the correct choice.

Comments

only leaf nodes are connected.

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yah i make a complete binary tree in which i connected all leaf node in series since its given they are tied together and then connect to battery and root to other side of battery.is this right?

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I would suppose it is $2^n$ resistors of resistance $n$ each, connected in parallel. None of the options matches. In TIFR marks were given for all for this question.

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ok!  each edge has one ohm resistor its given i considered for nodes.

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In this image leaf nodes D,E,F,G joined. Now make a single node by combining D,E,F,G and call it Z. (Because we have to specify two terminals/nodes between which we are calcutating resistance).

 

Between B and Z effective resistance will be 1/2. 

Between C and Z effective resistance will be 1/2.

Now AB(1 ohm) and BZ(1/2ohm) are in serial..and AC and CZ are also in serial.

AZ and AZ will be parallel(3/2 ohm each)

Net resistance between A and Z= 3/4..

 

If we calculate like this for a tree of height n then effective resistance will be

$\frac{2^n-1}{2^n}=O(1)$

@Arjun sir, please check once

 

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@Satbir 

@srestha please check above comment..

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TIFR2012-16

@Verma Ashish  @Arjun sir

Which answer is correct as TIFR is saying none of the option matches. But the best answer is concluding $'A'$ as the answer.

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@Kushagra गुप्ता i had already commented about how i interpreted this question..