**Question:**

(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.

**Solution:**

(a) Total number of resistors = *n*

Resistance of each resistor = *R*

(i) When *n* resistors are connected in series, effective resistance *R*1is the maximum, given by the product *nR*.

Hence, maximum resistance of the combination, *R*1 = *nR*

(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,

*R*1 = 1 Ω, *R*2 = 2 Ω, *R*3 = 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* + *R*

= 5 *R*

*= 2*

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.