# (a) Given n resistors each of resistance R, how will you combine them to get the

Question:

(a) Given 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.

Solution:

(a) Total number of resistors = n

Resistance of each resistor = R

(i) When n resistors are connected in series, effective resistance R1is the maximum, given by the product nR.

Hence, maximum resistance of the combination, RnR

(ii) When $n$ resistors are connected in parallel, the effective resistance $\left(R_{2}\right)$ is the minimum, given by the ratio $\frac{R}{n}$.

Hence, minimum resistance of the combination, $R_{2}=\frac{R}{n}$

(iii) The ratio of the maximum to the minimum resistance is,

$\frac{R_{1}}{R_{2}}=\frac{n R}{\frac{R}{n}}=n^{2}$

(b) The resistance of the given resistors is,

R1 = 1 Ω, R2 = 2 Ω, R3 = 3 Ω2

i. Equivalent resistance, $R^{\prime}=\frac{11}{3} \Omega$

Consider the following combination of the resistors.

Equivalent resistance of the circuit is given by,

$R^{\prime}=\frac{2 \times 1}{2+1}+3=\frac{2}{3}+3=\frac{11}{3} \Omega$

ii. Equivalent resistance, $R^{\prime}=\frac{11}{5} \Omega$

Consider the following combination of the resistors.

Equivalent resistance of the circuit is given by,

$R^{\prime}=\frac{2 \times 3}{2+3}+1=\frac{6}{5}+1=\frac{11}{5} \Omega$

(iii) Equivalent resistance, R = 6 Ω

Consider the series combination of the resistors, as shown in the given circuit.

Equivalent resistance of the circuit is given by the sum,

R = 1 + 2 + 3 = 6 Ω

(iv) Equivalent resistance, $R^{\prime}=\frac{6}{11} \Omega$

Consider the series combination of the resistors, as shown in the given circuit.

Equivalent resistance of the circuit is given by,

$R^{\prime}=\frac{1 \times 2 \times 3}{1 \times+2 \times 3+3 \times 1}=\frac{6}{11} \Omega$

(c) (a) It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in series.

Hence, their equivalent resistance = (1+1) = 2 Ω

It can also be observed that two resistors of resistance 2 Ω each are connected in series.

Hence, their equivalent resistance = (2 + 2) = 4 Ω.

Therefore, the circuit can be redrawn as

It can be observed that 2 Ω and 4 Ω resistors are connected in parallel in all the four loops. Hence, equivalent resistance (R) of each loop is given by,

$R=\frac{2 \times 4}{2+4}=\frac{8}{6}=\frac{4}{3} \Omega$

The circuit reduces to,

All the four resistors are connected in series.

Hence, equivalent resistance of the given circuit is $\frac{4}{3} \times 4=\frac{16}{3} \Omega$

(b) It can be observed from the given circuit that five resistors of resistance R each are connected in series.

Hence, equivalent resistance of the circuit = R + R + R + R

= 5 R

= 2