Answer –
(a) The total number of resistors is equal to n.
Each resistor’s resistance is equal to R.
(i) When the resistors are connected in series, the maximum effective resistance is obtained.
R1 = nR is the effective resistance
(ii) When the resistors are connected in parallel, the effective resistance is the lowest possible, given by R2 = R/n
nR/(R/n) = n2 is the ratio of greatest resistance to minimum resistance
(b) The resistance given are 1 Ω, 2 Ω, 3 Ω
(i) 11/3 Ω
Consider the following set of circumstances:
The resistors of resistance 1 Ω and 2 Ω are connected in parallel. Therefore, the effective resistance is given by –
\[\frac{1}{{{R}^{‘}}}=\left[ \frac{1}{1}+\frac{1}{2} \right]=\frac{3}{2}\]
So, R’ = 2/3 Ω
These resistors are then, connected in series with 3 Ω. Therefore the effective resistance is given by –
R = R’ + 3 = (2/3) +3
R = 11/3 Ω
(ii) 11/5 Ω Consider the following set of circumstances:
The resistors of resistance 2 Ω and 3 Ω are connected in parallel. Therefore, the effective resistance becomes –
\[\frac{1}{{{R}^{‘}}}=\left[ \frac{1}{3}+\frac{1}{2} \right]=\frac{5}{6}\]
R’ = 6/5 Ω
These resistors are then, connected in series with 3 Ω. Therefore the effective resistance is given by –
R = R’ + 3 = (6/5) +1
R = 11/5 Ω
(iii) 6 Ω
The effective resistance when the resistors with resistance 1 Ω, 2 Ω, 3 Ω are connected in series is = 1 Ω +2 Ω + 3 Ω = 6 Ω
(iv) (6/11) Ω
When the resistors are connected in parallel the effective resistance is
\[\frac{1}{{{R}^{‘}}}=\left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3} \right]=\frac{11}{6}\Omega \]
R’ = (6/11) Ω
(c) (a) Two resistors of resistance 1Ω are linked in series in each loop. As a result, (1+1)Ω = 2Ω is the effective resistance.
Similarly, two resistors of resistance 2Ω are linked in series in all loops. As a result, the effective resistance is equal to (2+2) Ω = 4Ω .
In all four loops, the 2Ω and 4Ω resistors are connected in parallel. As a result, effective resistance becomes –
\[\frac{1}{{{R}^{‘}}}=\left[ \frac{1}{2}+\frac{1}{4} \right]=\frac{3}{4}\Omega \]
R’ = 4/3 Ω
The four resistors are wired together in series. As a result, each comparable resistance R is (4/3)Ω × 4 = 16/3Ω
(c) (b) The resistors are wired in parallel. As a result, the effective resistance is R + R + R + R + R = 5R