Question

Question 3.20: (a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance? (b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω? (c) Determine the equivalent resistance of networks shown in Fig. 3.31.

Class 12 - Physics - Current Electricity Current Electricity Page-129

Answer 1

(a) (i)for getting maximum equivalent resistance , we should combine the resistors in series combination.
e.g., $R_{eq}=R_1+R_2+R_3+R_4+....$

n resistors each of resistance R are given
then, maximum equivalent resistance ,
Req = R + R + R + R + R +.... n times
Req = nR
hence, maximum equivalent resistance/effective resistance =nR

(ii) for getting minimum resistance , we should join the resistors in parallel combination.
e.g., $1/R_{eq}=1/R_1+1/R_2+1/R_3+....$

so, n resistors or resistance R are joined in parallel for minimum resistance .
e.g., 1/Req = 1/R + 1/R + 1/R + .... n times
Req = R/n

so, $\frac{R_{max}}{R_{min}}=n^2$

(b) see attachment ,

(c) (i) 1/Req = 1/(4/3) + 1/(4/3) + 1/(4/3) + 1/(4/3)
= 4/(4/3) = 16/3 ohm [ see figure ]

(ii) all the resistence are in series ,
so, Req = R + R + R + R + R = 5R