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.

3.20 Total number of resistors = n

Resistance of each resistor = R

When the resistors are connected in series, effective resistance $R_{eff}$ is maximum. $R_{eff}$ = nR

When n resistors are connected in parallel, the effective resistance, $R_{eff}$ is minimum. $R_{eff}$ = $\frac{R}{n}$ . The ratio of maximum to minimum resistance = $\frac{nR}{\frac{R}{n}}$ = $n^2$

Let us assume, $R_1$ = 1 Ω, $R_2$ = 2 Ω, $R_3$ = 3 Ω

Required equivalent resistance, R = $\frac{11}{3}$Ω

From the circuit the equivalent resistance is given by

R = $\frac{2\times 1}{2+1}$ + 3 = $\frac{2}{3}$ + 3 = $\frac{11}{3}$ Ω

Required equivalent resistance, R = $\frac{11}{5}$ Ω

F

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3.20 Total number of resistors = n

Resistance of each resistor = R

When the resistors are connected in series, effective resistance $R_{eff}$ is maximum. $R_{eff}$ = nR

When n resistors are connected in parallel, the effective resistance, $R_{eff}$ is minimum. $R_{eff}$ = $\frac{R}{n}$ . The ratio of maximum to minimum resistance = $\frac{nR}{\frac{R}{n}}$ = $n^2$

Let us assume, $R_1$ = 1 Ω, $R_2$ = 2 Ω, $R_3$ = 3 Ω

Required equivalent resistance, R = $\frac{11}{3}$Ω

From the circuit the equivalent resistance is given by

R = $\frac{2\times 1}{2+1}$ + 3 = $\frac{2}{3}$ + 3 = $\frac{11}{3}$ Ω

Required equivalent resistance, R = $\frac{11}{5}$ Ω

From the circuit, the equivalent resistance is given by

R = $\frac{2\times 3}{2+3}$ + 1 = $\frac{6}{5}$ + 1 = $\frac{11}{5}$ Ω

Required equivalent resistance, R = 6 Ω

From the circuit, the equivalent resistance is given by

R = 1 + 2+ 3 = 6 Ω

Required equivalent resistance, R = $\frac{6}{11}$ Ω

From the circuit, the equivalent resistance is given by

R = $\frac{1\times 2\times 3}{1\times 2+2\times 3+3\times 1}$ = $\frac{6}{11}$ Ω

(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 the series. Hence the equivalent resistance is = 1 + 1= 2 $\Omega$ . At the bottom part 2 resistors of 2 Ω each are connected in a series and thus make equivalent resistance of 2 + 2 = 4 Ω. Thus the circuit can be redrawn as 2 Ω and 4 Ω resistances are connected in parallel. Hence the equivalent resistance of each loop is R = $\frac{2\times 4}{2+4}$ = $\frac{4}{3}$ Ω. All these 4 loop resistors are connected in series. Thus the total equivalent resistance is $\frac{4}{3}$ Ω $\times 4=\frac{16}{3}$ Ω

Here all 5 resistors are connected in series. So the equivalent resistance is

5 $\times R=5R$

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<p><strong>3.20 </strong>Total number of resistors = n</p><p>Resistance of each resistor = R</p><p>When the resistors are connected in series, effective resistance&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mi>e</mi><mi>f</mi><mi>f</mi></mrow></mrow></msub></math></span>&nbsp;is maximum.&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mi>e</mi><mi>f</mi><mi>f</mi></mrow></mrow></msub></math></span>&nbsp;= nR</p><p>When n resistors are connected in parallel, the effective resistance,&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mi>e</mi><mi>f</mi><mi>f</mi></mrow></mrow></msub></math></span>&nbsp;is minimum.&nbsp;<span title="Click to copy mathml"><math><mi></mi><msub><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mi>e</mi><mi>f</mi><mi>f</mi></mrow></mrow></msub></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mi>n</mi></mrow></mrow></mfrac></math></span>&nbsp;. The ratio of maximum to minimum resistance =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>n</mi><mi>R</mi></mrow></mrow><mrow><mrow><mfrac><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mi>n</mi></mrow></mrow></mfrac></mrow></mrow></mfrac></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><msup><mrow><mrow><mi>n</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math></span></p><p>Let us assume,&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span>&nbsp;= 1 Ω,&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span>&nbsp;= 2 Ω,&nbsp;<span title="Click to copy mathml"><math><msub><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msub></math></span>&nbsp;= 3 Ω</p><p>Required equivalent resistance, R = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>11</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math></span>Ω</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728541143phpCFN8dA_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728541143phpCFN8dA.jpeg" alt="" width="287" height="69"></picture></div></div><p>From the circuit the equivalent resistance is given by</p><p>R =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2</mn><mo>×</mo><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn><mo>+</mo><mn>1</mn></mrow></mrow></mfrac></math></span>&nbsp;+ 3 =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math></span>&nbsp;+ 3 =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>11</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math></span>&nbsp;Ω</p><p>Required equivalent resistance, R =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>11</mn></mrow></mrow><mrow><mrow><mi></mi><mn>5</mn></mrow></mrow></mfrac></math></span>&nbsp;Ω</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728541166phpYc0bJw_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728541166phpYc0bJw.jpeg" alt="" width="257" height="60"></picture></div></div><p>From the circuit, the equivalent resistance is given by</p><p>R =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2</mn><mo>×</mo><mn>3</mn></mrow></mrow><mrow><mrow><mn>2</mn><mo>+</mo><mn>3</mn></mrow></mrow></mfrac></math></span>&nbsp;+ 1 =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>6</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></mfrac></math></span>&nbsp;+ 1 =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>11</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></mfrac></math></span>&nbsp;Ω</p><p>Required equivalent resistance, R = 6 Ω</p><p>From the circuit, the equivalent resistance is given by</p><p>R = 1 + 2+ 3 = 6 Ω</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728541191phpKg8YTV_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728541191phpKg8YTV.jpeg" alt="" width="310" height="58"></picture></div></div><p>Required equivalent resistance, R =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>6</mn></mrow></mrow><mrow><mrow><mn>11</mn></mrow></mrow></mfrac></math></span>&nbsp;Ω</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728541253phpySM29n_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728541253phpySM29n.jpeg" alt="" width="198" height="91"></picture></div></div><p>From the circuit, the equivalent resistance is given by</p><p>R =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn><mo>×</mo><mn>2</mn><mo>×</mo><mn>3</mn></mrow></mrow><mrow><mrow><mn>1</mn><mo>×</mo><mn>2</mn><mo>+</mo><mn>2</mn><mo>×</mo><mn>3</mn><mo>+</mo><mn>3</mn><mo>×</mo><mn>1</mn></mrow></mrow></mfrac></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>6</mn></mrow></mrow><mrow><mrow><mn>11</mn></mrow></mrow></mfrac></math></span>&nbsp;Ω</p><p>(a)</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728541277phpYOqMEp_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728541277phpYOqMEp.jpeg" alt="" width="280" height="70"></picture></div></div><p>It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in the series. Hence the equivalent resistance is = 1 + 1= 2&nbsp;<span title="Click to copy mathml"><math><mi>Ω</mi></math></span>&nbsp;. At the bottom part 2 resistors of 2 Ω each are connected in a series and thus make equivalent resistance of 2 + 2 = 4 Ω. Thus the circuit can be redrawn as 2 Ω and 4 Ω resistances are connected in parallel. Hence the equivalent resistance of each loop is R =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>2</mn><mi></mi><mo>×</mo><mn>4</mn></mrow></mrow><mrow><mrow><mn>2</mn><mo>+</mo><mn>4</mn></mrow></mrow></mfrac></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>4</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math></span>&nbsp;Ω. All these 4 loop resistors are connected in series. Thus the total equivalent resistance is&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>4</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac></math></span>&nbsp;Ω&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><mn>4</mn><mo>=</mo><mi></mi><mfrac><mrow><mrow><mn>16</mn></mrow></mrow><mrow><mrow><mn>3</mn><mi></mi></mrow></mrow></mfrac></math></span>&nbsp;Ω</p><p>Here all 5 resistors are connected in series. So the equivalent resistance is</p><p>5&nbsp;<span title="Click to copy mathml"><math><mo>×</mo><mi>R</mi><mo>=</mo><mn>5</mn><mi>R</mi></math></span></p>

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