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Let there be n resistors R1............Rn with Rmax = max (R1......... Rn) and Rmin = min {R1..... Rn}. Show that when they are connected in parallel, the resultant resistance RP < R min - Physics

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Question

Let there be n resistors R1............Rn with Rmax = max (R1......... Rn) and Rmin = min {R1..... Rn}. Show that when they are connected in parallel, the resultant resistance RP <  R min and when they are connected in series, the resultant resistance RS > Rmax. Interpret the result physically.

Short Note

Solution

Parallel grouping: Same potential difference appeared across each resistance but current distributes in the reverse ratio of their resistance, i.e. `i  oo 1/R`


Series grouping: Same current flows through each resistance but potential difference distributes in the ratio of resistance, i.e. `V  oo  R`

In parallel combination: When all resistances are connected in parallel, the equivalent resistance Rρ is given by

`1/R_ρ = 1/R_1 + ... + 1/R_n`

By multiplying both sides by Rmin, we have

`R_(min)/R_ρ = R_(min)/R_1 + R_(min)/R_2 + ... + R_(min)/R_n`

Here, in RHS, there exists one term `R_(min)/R_(min)` = 1 and other terms are positive, so we have

`R_(min)/R_ρ = R_(min)/R_1 + R_(min)/R_2 + ... + R_(min)/R_n > 1`

This shows that the resultant resistance Rρ < Rmin.

Thus, in parallel combination, the equivalent resistance of resistors is even less than the minimum resistance available in a combination of resistors.

In series combination: When all resistances are connected in series, the equivalent resistance Rs is given by

Rs = R1 + ... + Rn

Here, in RHS, there exist one term having resistance Rmax.

So, we have

or Rs = R1 + ... + Rmax ... + ... + Rn

Rs = R1 + ... + Rmax ... + Rn = Rmax + ... (R1 + ... + )Rn

or Rs ≥ Rmax

Rs = Rmax(R1 + ... + Rn)

Thus, in series combination, the equivalent resistance of resistors is greater than the maximum resistance available in a combination of resistors.

Physical interpretation:


(a)

(b)

In figure (b), Rmin provides an equivalent routine as in figure. (a) for current. But in addition, there are (n – 1) routes by the remaining (n – 1) resistors. Current in figure. (b) is greater than current in figure (a) Effective resistance in figure. (b) < Rmin. Second circuit evidently affords a greater resistance.


(c)

(d)

In figure (d), Rmax provides an equivalent route as in figure. (c)  for current. Current in figure (d) < current in figure (c). Effective resistance in figure. (d) > Rmax. Second circuit evidently affords a greater resistance.

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Chapter 3: Current Electricity - MCQ I [Page 20]

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NCERT Exemplar Physics [English] Class 12
Chapter 3 Current Electricity
MCQ I | Q 3.23 | Page 20

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