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Question

A network of resistances is constructed with R1 and R2 as shown in the figure. The potential at the points 1, 2, 3 … n are V1,V2,V3,Vn respectively, each having a potential k times smaller than the previous one.

The ratio R2R3 is

A
(k1)2k
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B
k21k
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C
kk1
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D
k1k2
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Solution

The correct option is C kk1
Since each point have potential k times smaller than previous one, thus V1=Vok, V2=V1k, V3=V2k and so on. Let current I flow through R1 and gets distributed as shown.

Current ,I=I1+I2(1)
Since current, I=Potential differenceresistance
Therefore equation (1) can be written as,
VoV1R1=V1V2R1+Vo0R2(2)
Substituting V1=Vok, V2=V1k=Vok2 in (2),
V2k2V2kR1=V2kV2R1+V2kR2
k2kR1=k1R1+kR2
On rearranging the above relation, we get
R1R2=(k1)2k
R1=R2(k1)2k(3)
Consider the last branch of the circuit and let current flow through R1 as shown.
Now current through resistor R1 of the last branch in the circuit and R3 is same.
Therefore equating currents in form of potential difference and resistance,
Vn1VnR1=Vn0R3
R1R3=Vn1Vn1
Substituting the value of R1 from (3),
R2(k1)2kR3=Vn1Vn1(4)
Since it is given that potential at a point is k times smaller than previous one, therefore Vn=Vn1k
Therefore (4) becomes,
R2(k1)2kR3=k1
R2R3=kk1

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