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
(k−1)2k
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B
k2−1k
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C
kk−1
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D
k−1k2
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Solution
The correct option is Ckk−1 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, Vo−V1R1=V1−V2R1+Vo−0R2(2)
Substituting V1=Vok, V2=V1k=Vok2 in (2), V2k2−V2kR1=V2k−V2R1+V2kR2 k2−kR1=k−1R1+kR2
On rearranging the above relation, we get R1R2=(k−1)2k ⇒R1=R2(k−1)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, Vn−1−VnR1=Vn−0R3 R1R3=Vn−1Vn−1
Substituting the value of R1 from (3), R2(k−1)2kR3=Vn−1Vn−1(4)
Since it is given that potential at a point is k times smaller than previous one, therefore Vn=Vn−1k
Therefore (4) becomes, R2(k−1)2kR3=k−1 ⇒R2R3=kk−1