Resisting resistance

Electricity and Magnetism Level 3

In the given diagram, if the equivalent resistance between the points A and B can be written as R ′ = a R + b r c + d R r R' = \frac{aR + br}{c + \frac{dR}r} . What is a + b + c + d a+b+c+d ?

Note that the expression is in its simplest form.


The answer is 8.

Somsubhra Ghosh by Somsubhra Ghosh , 25, India

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2 solutions

Chew-Seong Cheong
Jun 27, 2015

Transform the left (A-side) delta connection into a star connection, then we have:

R ′ = R r R + 2 r + ( r 2 R + 2 r + R ) ∣ ∣ ( R r R + 2 r + r ) = R r R + 2 r + r 2 + 2 R r + R 2 R + 2 r ∣ ∣ R r + R r + 2 r 2 R + 2 r = R r R + 2 r + ( R + r ) 2 R + 2 r ∣ ∣ 2 r ( R + r ) R + 2 r = R r R + 2 r + ( R + r ) 2 R + 2 r × 2 r ( R + r ) R + 2 r ( R + r ) 2 R + 2 r + 2 r ( R + r ) R + 2 r = R r R + 2 r + 2 r ( R + r ) 3 ( R + 2 r ) 2 ( R + r ) ( R + 3 r ) R + 2 r = R r R + 2 r + 2 r ( R + r ) 2 ( R + 2 r ) ( R + 3 r ) = r R + 2 r ( R + 2 ( R + r ) 2 R + 3 r ) = r R + 2 r ( R 2 + 3 R r + 2 R 2 + 4 R r + 2 r 2 R + 3 r ) = r ( 3 R 2 + 7 R r + 2 r 2 ) ( R + 2 ) ( R + 3 r ) = r ( 3 R + r ) ( R + 2 r ) ( R + 2 ) ( R + 3 r ) = r ( 3 R + r ) R + 3 r = 3 R + r 3 + R r \begin{aligned} R' & = \frac{Rr}{R+2r} + \left(\frac{r^2}{R+2r}+R\right) || \left(\frac{Rr}{R+2r}+r \right) \\ & = \frac{Rr}{R+2r} + \frac{r^2+2Rr+R^2}{R+2r} || \frac{Rr+Rr+2r^2}{R+2r} \\ & = \frac{Rr}{R+2r} + \frac{(R+r)^2}{R+2r} || \frac{2r(R+r)}{R+2r} \\ & = \frac{Rr}{R+2r} + \frac {\frac{(R+r)^2}{R+2r} \times \frac{2r(R+r)}{R+2r}} {\frac{(R+r)^2}{R+2r} + \frac{2r(R+r)}{R+2r}} \\ & = \frac{Rr}{R+2r} + \frac {\frac{2r(R+r)^3}{(R+2r)^2}} {\frac{(R+r)(R+3r)}{R+2r}} \\ & = \frac{Rr}{R+2r} + \frac {2r(R+r)^2}{(R+2r)(R+3r)} \\ & = \frac{r}{R+2r}\left(R + \frac {2(R+r)^2}{R+3r} \right) \\ & = \frac{r}{R+2r}\left(\frac {R^2+3Rr + 2R^2+4Rr+2r^2}{R+3r} \right) \\ & = \frac {r(3R^2+7Rr+2r^2)}{(R+2)(R+3r)} = \frac {r(3R+r)(R+2r)}{(R+2)(R+3r)} \\ & = \frac {r(3R+r)}{R+3r} = \frac {3R+r}{3+\frac{R}{r}} \end{aligned}

⇒ a + b + c + d = 3 + 1 + 3 + 1 = 8 \Rightarrow a + b + c + d = 3+1+3+1 = \boxed{8}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

You are a genius basher sir ! Your latex skills are amazing too !

Venkata Karthik Bandaru - 5 years, 11 months ago
Somsubhra Ghosh
Jun 24, 2015

This is a simple case of an unbalanced Wheatstone bridge. To find the equivalent resistance of the circuit, let us connect a battery of pd, V betn the points A and B as shown in the figure.

Applying Kirchhoff's voltage rule to loop 1, -(I - I')r - I"r + I'R = 0

For loop 2, -(I - I' - I")R + (I' + I")r + I"r =0

For loop 3, V - I'R - (I' + I")r =0

Solving the three equations, we get I = V (3+ R/r) / (3R + r)

Comparing this with I = V/R', we get R' = (3R + r)/(3+ R/r) Hence the result.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Don't mind but have u given it from irodov??

Shyambhu Mukherjee - 5 years, 6 months ago

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You can find the same problem with capacitors in the worked out examples of H.C. Verma.

Soumava Pal - 5 years, 2 months ago

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