Class 10 Chapter 1 Electricity Physics Solution S.Chand. Part 5

Part 5

COMBINATION OF RESISTANCES (OR RESISTORS)

Very Short Answer Type Questions

1. The law of combination of resistances in series states that the total resistance

�_{�}

of resistors connected in series is equal to the sum of their individual resistances. Mathematically, it can be represented as:

�_{�} = �_{1} + �_{2} + �_{3} + \dots

2. If five resistances, each of value 0.2 ohm, are connected in series, the resultant resistance (

�_{�}

) will be:

�_{�} = 5 \times 0.2 = 1 Ω

3. The law of combination of resistances in parallel states that the reciprocal of the total resistance (

�_{�}

) of resistors connected in parallel is equal to the sum of the reciprocals of their individual resistances. Mathematically, it can be represented as:

\frac{1}{�_{�}} = \frac{1}{�_{1}} + \frac{1}{�_{2}} + \frac{1}{�_{3}} + \dots

4. If 3 resistances of 3 ohms each are connected in parallel, their total resistance (

�_{�}

) will be:

\frac{1}{�_{�}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 Ω

�_{�} = 1 Ω

5. To produce an equivalent resistance of 1 ohm using two resistances of 2 ohms each, they should be connected in parallel.

(i) When X and Y are connected in parallel, the resultant resistance will be less than either of the individual resistances.
(ii) When X and Y are connected in series, the resultant resistance will also be less than either of the individual resistances.

7. The possible values of resultant resistance obtained by combining two resistances, one of value 2 ohms and the other 6 ohms, are:

When connected in series: $2 + 6 = 8 Ω$
When connected in parallel: $\frac{2 \times 6}{2 + 6} = 1.5 Ω$

(a) To produce a combined resistance of 2 ohms, the two 4 ohm resistors should be connected in parallel.
(b) To produce a combined resistance of 8 ohms, the two 4 ohm resistors should be connected in series.

9. A have 10 ohm resistance but B have 0.101 ohm resistance so B have lower resistance

10.When a wire with resistance

$�$ is cut into two equal pieces and joined in parallel, the equivalent resistance $�_{eq}$ can be found using the formula for resistors in parallel:

$\frac{1}{�_{eq}} = \frac{1}{�_{1}} + \frac{1}{�_{2}}$

Since the two pieces are of equal resistance, let's call the resistance of each piece $�_{1}$ and $�_{2}$ . So, $�_{1} = �_{2} = \frac{�}{2}$ .

Now, substituting $�_{1} = �_{2} = \frac{�}{2}$ into the formula:

$\frac{1}{�_{eq}} = \frac{1}{\frac{�}{2}} + \frac{1}{\frac{�}{2}}$

Simplify:

$\frac{1}{�_{eq}} = \frac{2}{�}$

$�_{eq} = \frac{�}{2}$

So, when the wire is cut into two equal pieces and joined in parallel, the resistance of the combination becomes half of the original resistance $�$ , which is $�_{eq} = \frac{�}{2}$ .

11. (a) To find the combined resistance of resistors

$�_{1} = 500 Ω$ and $�_{2} = 1000 Ω$ connected in series, you simply add their resistances together because resistors in series add up:

$�_{total} = �_{1} + �_{2}$

$�_{total} = 500 Ω + 1000 Ω$

$�_{total} = 1500 Ω$

So, the combined resistance of the two resistors connected in series is $1500 Ω$

1500 Ω

. (b) For two resistors

$�_{1}$ and $�_{2}$ with the same resistance $2 Ω$ :

$�_{1} = �_{2} = 2 Ω$

Substitute the values into the formula:

$\frac{1}{�_{total}} = \frac{1}{2 Ω} + \frac{1}{2 Ω}$

$\frac{1}{�_{total}} = \frac{1}{2} + \frac{1}{2}$

$\frac{1}{�_{total}} = \frac{1}{2} + \frac{1}{2} = 1$

To find $�_{total}$ , take the reciprocal of both sides:

$�_{total} = \frac{1}{1}$

$�_{total} = 1 Ω$

So, the combined resistance of $2 Ω$ and $2 Ω$ resistors in parallel is $1 Ω$ .

(c) To find the combined resistance of the circuit, first, we need to calculate the equivalent resistance of the resistors in parallel, and then we'll add the series resistor.

For resistors in parallel, the formula to find the combined resistance is:

$\frac{1}{�_{parallel}} = \frac{1}{�_{1}} + \frac{1}{�_{2}}$

Given that both $�_{1}$ and $�_{2}$ are $4$ ohms:

$\frac{1}{�_{parallel}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

So, the equivalent resistance for the parallel combination is $�_{parallel} = 2$ ohms.

Now, we add this equivalent resistance to the series resistor $�_{3} = 3$ ohms:

$�_{total} = �_{parallel} + �_{3} = 2 + 3 = 5 Ω$

Therefore, the combined resistance of the circuit is $5$ ohms.

12. To find the current flowing through each resistor

$�_{1}$ and $�_{2}$ , we can use Ohm's Law again:

For resistor $�_{1}$ : $�_{1} = \frac{�}{�_{1}}$
For resistor $�_{2}$ : $�_{2} = \frac{�}{�_{2}}$

Given that $� = 24$ volts, $�_{1} = 6$ ohms, and $�_{2} = 4$ ohms:

For $�_{1}$ : $�_{1} = \frac{24 V}{6 Ω} = 4 A$
For $�_{2}$ : $�_{2} = \frac{24 V}{4 Ω} = 6 A$

So, the current flowing through $�_{1}$ is $4$ Amperes, and the current flowing through $�_{2}$ is $6$ Amperes.

Short Answer Type Questions

13. (a) Series Combination of Resistances: In a series combination, resistors are connected end-to-end so that the same current flows through each resistor. The total resistance is the sum of the individual resistances. Diagrammatically, it can be represented as follows:

------[R1]------[R2]------[R3]------ (Series Combination)

(b) Parallel Combination of Resistances: In a parallel combination, resistors are connected across the same two points, allowing multiple paths for the current to flow. The reciprocal of the total resistance is the sum of the reciprocals of the individual resistances. Diagrammatically, it can be represented as follows:

|--[R1]--|

| |

----| |---- (Parallel Combination)

| |

|--[R2]--|

(i) The resultant resistance is less than either of the individual resistances in a parallel combination. (ii) The resultant resistance is more than either of the individual resistances in a series combination.

14.

Given:

Battery voltage (V) = 9 V
Resistances: 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω, and 12 Ω To find the current through the 12 Ω resistor, we first need to calculate the total resistance of the circuit using the series combination formula $�_{total} = �_{1} + �_{2} + �_{3} + �_{4} + �_{5}$ , and then use Ohm's Law $� = \frac{�}{�}$ .

15. Circuit Diagram:

Battery (6 V) ---[20 Ω bulb]---[4 Ω resistance wire]---

(a) Total resistance of the circuit ( $�_{total}$ ): $�_{total} = 20 Ω + 4 Ω = 24 Ω$

(b) Current through the circuit ( $�$ ) using Ohm's Law ( $� = \frac{�}{�}$ ): $� = \frac{6 V}{24 Ω} = 0.25 A$

(c) Potential difference across the electric bulb ( $�_{bulb}$ ) using Ohm's Law ( $� = � \times �$ ): $�_{bulb} = 0.25 A \times 20 Ω = 5 V$

(d) Potential difference across the resistance wire ( $�_{wire}$ ): $�_{wire} = 0.25 A \times 4 Ω = 1 V$

16. (i) Given that the two resistors (10 ohm and 15 ohm) are connected in parallel and a current of 1 ampere is flowing through a separate resistor of 5 ohms, we can use the fact that the total current in a series circuit is equal at any point.

Since the 5 ohm resistor is in series with the parallel combination of the 10 ohm and 15 ohm resistors, the total current flowing through the circuit is 1 ampere.

Now, for the resistors in parallel, the total current splits between them according to their resistances. The current through each resistor can be calculated using Ohm's Law:

For the parallel combination of 10 ohm and 15 ohm resistors: $�_{parallel} = \frac{1}{\frac{1}{�_{1}} + \frac{1}{�_{2}}}$ $�_{parallel} = \frac{1}{\frac{1}{10} + \frac{1}{15}} = \frac{1}{\frac{3}{30} + \frac{2}{30}} = \frac{1}{\frac{5}{30}} = \frac{30}{5} = 6 Ω$

Now, using Ohm's Law, we can find the current through each resistor in parallel: $� = \frac{�}{�}$

For the 10 ohm resistor: $�_{10} = \frac{�}{10}$

For the 15 ohm resistor: $�_{15} = \frac{�}{15}$

Since the potential difference across both resistors in parallel is the same, we can say: $�_{10} = �_{15}$

Given that the total current in the circuit is 1 ampere, and the total resistance of the parallel combination is 6 ohms, we can find the potential difference across the parallel combination: $� = � \times � = 1 A \times 6 Ω = 6 V$

Now, we can find the current through each resistor in parallel: $�_{10} = \frac{6 V}{10 Ω} = 0.6 A$ $�_{15} = \frac{6 V}{15 Ω} = 0.4 A$

Therefore, the current through the 10 ohm resistor is $0.6 A$ and through the 15 ohm resistor is $0.4 A$

$0.4 A$ .

(ii) Potential difference across AB (5 ohm resistor):

Since the current flowing through the 5 ohm resistor is 1 ampere, we can use Ohm's Law directly: $�_{� �} = � \times �_{� �} = 1 A \times 5 Ω = 5 V$

Therefore, the potential difference across AB is $5 V$

$5 V$ .

Total potential difference across AC:

Since the 5 ohm resistor and the parallel combination of 10 ohm and 15 ohm resistors are in series, the total potential difference across AC is the sum of the potential differences across each segment: $�_{� �} = �_{� �} + �_{parallel} = 5 V + 6 V = 11 V$

=5V+6V=11V

(iii) The total resistance of the circuit can be calculated by considering the resistors in series and the parallel combination.

Resistors in series: The 5 ohm resistor (AB) and the parallel combination of the 10 ohm and 15 ohm resistors (BC) are in series. Therefore, the total resistance due to the series combination is the sum of the resistances of these two segments: $�_{series} = �_{� �} + �_{parallel} = 5 Ω + 6 Ω = 11 Ω$
Total resistance: The total resistance of the circuit is the total resistance due to the series combination, which we found to be $11 Ω$ .

So, the total resistance of the circuit is $11 Ω$ .

17. (i) 0.44 A (ii) 3.2 V

18.To obtain minimum and maximum currents in the circuit with two resistors of 5 Ω and 10 Ω connected to a battery of 6 V, we need to consider the scenarios where resistors are connected in series and parallel.

(a) (i) To obtain minimum current flowing, the resistors should be connected in series. (ii) To obtain maximum current flowing, the resistors should be connected in parallel.

(b) Let's calculate the strength of the total current in each case:

(i) Minimum current flowing (Resistors in series): Total resistance ( $�_{total}$ ) in series = $�_{1} + �_{2}$ = 5 Ω + 10 Ω = 15 Ω Using Ohm's law, $� = � �$ , where $�$ is the voltage (6 V) and $�$ is the resistance, we can find the current ( $�_{min}$ ) flowing: $�_{min} = \frac{�}{�_{total}} = \frac{6 V}{15 Ω} = 0.4 A$

(ii) Maximum current flowing (Resistors in parallel): Total resistance ( $�_{total}$ ) in parallel = $\frac{�_{1} \times �_{2}}{�_{1} + �_{2}}$ = $\frac{5 \times 10}{5 + 10}$ = $\frac{50}{15}$ = $\frac{10}{3}$ Ω Using Ohm's law, $� = � �$ , we can find the current ( $�_{max}$ ) flowing: $�_{max} = \frac{�}{�_{total}} = \frac{6 V}{\frac{10}{3} Ω} = \frac{18}{10} A = 1.8 A$

So, to summarize: (i) Minimum current flowing: Connect the resistors in series. The strength of the total current in the circuit is 0.4 A. (ii) Maximum current flowing: Connect the resistors in parallel. The strength of the total current in the circuit is 1.8 A.

19.

20.

21.

To determine how many resistors with a resistance of 176 ohms should be connected in parallel to draw a current of 5 amperes from a 220-volt supply line, we need to use Ohm's law ( $� = � �$ ) and the formula for resistors connected in parallel.

Given:

Supply voltage ( $�$ ) = 220 V
Total current ( $�$ ) = 5 A
Resistance of one resistor ( $�_{resistor}$ ) = 176 Ω

Let $�$ be the number of resistors connected in parallel.

Using Ohm's law, we can find the total resistance ( $�_{total}$ ) required: $�_{total} = \frac{�}{�} = \frac{220 V}{5 A} = 44 Ω$

Now, for resistors connected in parallel, the total resistance ( $�_{total}$ ) is given by: $\frac{1}{�_{total}} = \frac{1}{�_{resistor}} + \frac{1}{�_{resistor}} + \dots + \frac{1}{�_{resistor}}$ $\frac{1}{44} = \frac{�}{176}$ $� = \frac{176}{44} = 4$

So, 4 resistors with a resistance of 176 ohms each should be connected in parallel to draw a current of 5 amperes from a 220-volt supply line.

22.

Given:

Supply voltage ( $�$ ) = 220 V
Resistance of coil A ( $�_{�}$ ) = 24 Ω
Resistance of coil B ( $�_{�}$ ) = 24 Ω

(a) When only one coil A is used: Using Ohm's law ( $� = � �$ ), we can find the current ( $�$ ) drawn: $� = \frac{�}{�_{�}} = \frac{220 V}{24 Ω} = 9.166 A$

(b) When coils A and B are used in series: Total resistance ( $�_{total}$ ) in series = $�_{�} + �_{�}$ = $24 Ω + 24 Ω = 48 Ω$ Using Ohm's law, we can find the current ( $�$ ) drawn: $� = \frac{�}{�_{total}} = \frac{220 V}{48 Ω} = 4.583 A$

(c) When coils A and B are used in parallel: Total resistance ( $�_{total}$ ) in parallel = $\frac{�_{�} \times �_{�}}{�_{�} + �_{�}}$ = $\frac{24 \times 24}{24 + 24} = 12 Ω$ Using Ohm's law, we can find the current ( $�$ ) drawn: $� = \frac{�}{�_{total}} = \frac{220 V}{12 Ω} = 18.333 A$

So, to summarize: (a) When only one coil A is used, the current drawn is approximately 9.166 A. (b) When coils A and B are used in series, the current drawn is approximately 4.583 A. (c) When coils A and B are used in parallel, the current drawn is approximately 18.333 A.

23.

24.

25.Given:

Potential difference ( $�$ ) = 4 V
Resistance of the first resistor ( $�_{1}$ ) = 6 Ω
Resistance of the second resistor ( $�_{2}$ ) = 2 Ω

(a) Combined resistance ( $�_{total}$ ) in series: $�_{total} = �_{1} + �_{2} = 6 Ω + 2 Ω = 8 Ω$

(b) To find the current flowing, we'll use Ohm's Law, $� = � �$ , where $�$ is the current flowing: $� = \frac{�}{�_{total}} = \frac{4 V}{8 Ω} = 0.5 A$

So, to summarize: (a) The combined resistance is $8 Ω$ . (b) The current flowing through the circuit is $0.5 A$ . (c) The potential difference across the $6 Ω$ resistor is $3 V$ .

26.Given:

Potential difference ( $�$ ) = 6 V
Resistance of the first resistor ( $�_{1}$ ) = 3 Ω
Resistance of the second resistor ( $�_{2}$ ) = 6 Ω

(a) Combined resistance ( $�_{total}$ ) in parallel: $\frac{1}{�_{total}} = \frac{1}{�_{1}} + \frac{1}{�_{2}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$ $�_{total} = \frac{1}{\frac{1}{2}} = 2 Ω$

(b) Current flowing in the main circuit: $� = \frac{�}{�_{total}} = \frac{6 V}{2 Ω} = 3 A$

(c) Current flowing in the $3 Ω$ resistor ( $�_{3 Ω}$ ): Since the resistors are in parallel, the potential difference across each resistor is the same. Using Ohm's Law, we can find the current flowing through the $3 Ω$ resistor: $�_{3 Ω} = \frac{�}{�_{1}} = \frac{6 V}{3 Ω} = 2 A$

So, to summarize: (a) The combined resistance is $2 Ω$ . (b) The current flowing in the main circuit is $3 A$ . (c) The current flowing in the $3 Ω$ resistor is $2 A$ .

27.

28.

29.(a) Circuit diagram:

----[20Ω]----[20Ω]---- | | ----------------------- Battery (+) -----> Battery (-)

The arrow indicates the direction of conventional current flow, which is from the positive terminal of the battery to the negative terminal.

(b) Effective resistance ( $�_{total}$ ) of the two resistors in series: $�_{total} = �_{1} + �_{2} = 20 Ω + 20 Ω = 40 Ω$

(c) To calculate the current ( $�$ ) flowing from the battery, we'll use Ohm's Law ( $� = � �$ ), where $�$ is the potential difference (5 V) and $�_{total}$ is the effective resistance: $� = \frac{�}{�_{total}} = \frac{5 V}{40 Ω} = 0.125 A$

(d) Potential difference ( $�_{resistor}$ ) across each resistor: Since the resistors are in series, the potential difference across each resistor is the same and equal to the potential difference of the battery (5 V). So, the potential difference across each resistor is 5 V.

So, to summarize: (a) Circuit diagram drawn. (b) The effective resistance of the two resistors is $40 Ω$ . (c) The current that flows from the battery is $0.125 A$ . (d) The potential difference across each resistor is $5 V$ .

30.

31. Given:

Resistance of the first coil ( $�_{1}$ ) = 4 Ω
Resistance of the second coil ( $�_{2}$ ) = 2 Ω
Total current passing through the coils ( $�_{total}$ ) = 3 A

To find: (a) Combined resistance ( $�_{total}$ ) of the coils in parallel. (b) Current passing through the 2 Ω coil ( $�_{2 Ω}$ ).

(a) Combined resistance ( $�_{total}$ ) of the coils in parallel: $\frac{1}{�_{total}} = \frac{1}{�_{1}} + \frac{1}{�_{2}} = \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}$ $�_{total} = \frac{4}{3} Ω$

(b) Current passing through the 2 Ω coil ( $�_{2 Ω}$ ): Since the resistors are in parallel, the potential difference across each resistor is the same. Using Ohm's Law, we can find the current flowing through the 2 Ω coil: $�_{2 Ω} = �_{total} - �_{4 Ω}$ $�_{2 Ω} = 3 A - �_{4 Ω}$

We know that in a parallel circuit, the currents through each resistor add up to the total current ( $�_{total}$ ). So, the current passing through the 4 Ω coil ( $�_{4 Ω}$ ) is: $�_{4 Ω} = �_{total} - �_{2 Ω}$ $�_{4 Ω} = 3 A - �_{2 Ω}$

Now, let's substitute $�_{total}$ and $�_{4 Ω}$ to find $�_{2 Ω}$ : $\frac{�}{�_{total}} = 3 A - �_{2 Ω}$ $\frac{�}{\frac{4}{3} Ω} = 3 A - �_{2 Ω}$ $\frac{3 �}{4} = 3 A - �_{2 Ω}$ $\frac{3 \times 4}{4} = 3 A - �_{2 Ω}$ $3 = 3 A - �_{2 Ω}$ $�_{2 Ω} = 3 A - 3 A$ $�_{2 Ω} = 0 A$

So, to summarize: (a) The combined resistance of the coils in parallel is $\frac{4}{3} Ω$ . (b) The current passing through the $2 Ω$ coil is $0 A$ .

Long Answer Type Questions

32.

33.

34.

35.

Multiple Choice Questions (MCQs)

36.

37.

38.

39.

40.

41.

42.

Questions Based on High Order Thinking Skills (HOTS)

43.To achieve the desired combinations of resistances using three resistors each of 6 Ω, we need to connect them in different configurations.

(i) To obtain a combination with a resistance of 9 Ω:

Connect the three resistors in series. The total resistance in a series circuit is the sum of the individual resistances.

----[6Ω]----[6Ω]----[6Ω]---- Total Resistance = 18Ω

(ii) To obtain a combination with a resistance of 4 Ω:

Connect two resistors in parallel and then connect the third resistor in series with the parallel combination.

----[6Ω]----[6Ω]---- | | ----[6Ω]----[6Ω]---- Total Resistance = 4Ω

In this configuration, the parallel combination of two 6 Ω resistors has an equivalent resistance of 3 Ω. When this parallel combination is connected in series with the third 6 Ω resistor, the total resistance becomes 4 Ω.

44.

Let's denote the resistances as $�_{1}$ and $�_{2}$ .

Given:

Resultant resistance in parallel ( $�_{parallel}$ ) = 2 Ω
Resultant resistance in series ( $�_{series}$ ) = 9 Ω

We know that the formula for resistors in parallel is: $\frac{1}{�_{parallel}} = \frac{1}{�_{1}} + \frac{1}{�_{2}}$

Given $�_{parallel} = 2$ Ω, we can find $�_{2}$ in terms of $�_{1}$ .

We also know that the formula for resistors in series is simply: $�_{series} = �_{1} + �_{2}$

By substituting the expression for $�_{2}$ from the parallel connection formula into the series connection formula, we can solve for $�_{1}$ and $�_{2}$ .

Let's solve it step by step.

From the parallel connection formula: $\frac{1}{�_{parallel}} = \frac{1}{�_{1}} + \frac{1}{�_{2}}$ $\frac{1}{2} = \frac{1}{�_{1}} + \frac{1}{�_{2}}$

Since $�_{2} = \frac{�_{1} \cdot �_{parallel}}{�_{1} - �_{parallel}}$ , we substitute this expression into the series connection formula:

$�_{series} = �_{1} + \frac{�_{1} \cdot �_{parallel}}{�_{1} - �_{parallel}}$

Given $�_{series} = 9$ Ω, we can solve for $�_{1}$ .

$9 = �_{1} + \frac{�_{1} \cdot 2}{�_{1} - 2}$

$9 = �_{1} + \frac{2 �_{1}}{�_{1} - 2}$

$9 (�_{1} - 2) = �_{1} (�_{1} - 2) + 2 �_{1}$

$9 �_{1} - 18 = �_{1}^{2} - 2 �_{1} + 2 �_{1}$

$0 = �_{1}^{2} - 11 �_{1} + 18$

Using quadratic formula: $�_{1} = \frac{- � \pm \sqrt{�^{2} - 4 � �}}{2 �}$ where $� = 1$ , $� = - 11$ , and $� = 18$ .

$�_{1} = \frac{- (- 11) \pm \sqrt{(- 11)^{2} - 4 \cdot 1 \cdot 18}}{2 \cdot 1}$

$�_{1} = \frac{11 \pm \sqrt{121 - 72}}{2}$

$�_{1} = \frac{11 \pm \sqrt{49}}{2}$

$�_{1} = \frac{11 \pm 7}{2}$

Two possible solutions: $�_{1} = \frac{11 + 7}{2} = 9 Ω$ $�_{1} = \frac{11 - 7}{2} = 2 Ω$

Since we know that $�_{1}$ and $�_{2}$ cannot be the same (otherwise, it would lead to a zero resultant resistance in series), the correct value for $�_{1}$ is $9 Ω$ . Then, we can find $�_{2}$ using the expression we derived earlier:

$�_{2} = \frac{�_{1} \cdot �_{parallel}}{�_{1} - �_{parallel}}$

$�_{2} = \frac{9 \cdot 2}{9 - 2} = \frac{18}{7} \approx 2.571 Ω$

So, the values of the resistances are approximately $9 Ω$ and $2.571 Ω$ .

45.When resistors are connected in parallel, the formula for calculating the total resistance (

$�_{total}$ ) is:

$\frac{1}{�_{total}} = \frac{1}{�_{1}} + \frac{1}{�_{2}}$

Given:

Resistance of the first resistor ( $�_{1}$ ) = 8 ohms
Resultant resistance of the combination ( $�_{total}$ ) = 4.8 ohms

We need to find the value of the resistor $�$ , which we'll denote as $�_{2}$ .

Substituting the given values into the formula for resistors in parallel, we have:

$\frac{1}{4.8} = \frac{1}{8} + \frac{1}{�_{2}}$

Now, let's solve for $�_{2}$ :

$\frac{1}{�_{2}} = \frac{1}{4.8} - \frac{1}{8}$

$\frac{1}{�_{2}} = \frac{5}{24} - \frac{3}{24}$

$\frac{1}{�_{2}} = \frac{2}{24}$

$\frac{1}{�_{2}} = \frac{1}{12}$

$�_{2} = 12 ohms$

So, the value of the resistor $�$ is $12 ohms$ .

46.To obtain the desired resistances using resistors of 1, 2, and 3 ohms, we can combine them in various configurations. Let's illustrate each case:

(i) To obtain a resistance of 6 Ω:

Connect the resistors of 2 Ω and 3 Ω in series, then connect this series combination in parallel with the 1 Ω resistor.

----[1Ω]---- ----[2Ω]----[3Ω]---- | | ---------------------------------------- Total Resistance = 6Ω

(ii) To obtain a resistance of 11 Ω:

Connect the resistors of 1 Ω and 3 Ω in series, then connect this series combination in parallel with the 2 Ω resistor.

----[2Ω]---- ----[1Ω]----[3Ω]---- | | ---------------------------------------- Total Resistance = 11Ω

(iii) To obtain a resistance of 1.5 Ω:

Connect the 1 Ω and 2 Ω resistors in parallel, then connect this parallel combination in series with the 3 Ω resistor.

----[1Ω]----[2Ω]---- | | -----------------------[3Ω]---- Total Resistance = 1.5Ω

These configurations illustrate how you can achieve the desired resistances using the given resistors.

47.To obtain a resultant resistance of 2.5 Ω using resistors of 2 Ω, 3 Ω, and 5 Ω, we need to combine them in a specific configuration. Here's how we can do it:

Connect the 2 Ω resistor in series with the 3 Ω resistor, then connect this series combination in parallel with the 5 Ω resistor.

----[2Ω]----[3Ω]---- | | -----------------------[5Ω]---- Total Resistance = 2.5Ω

This configuration ensures that the overall resistance is 2.5 Ω.

48.To obtain total resistances of 4 Ω and 1 Ω using resistors of 2 Ω, 3 Ω, and 6 Ω, we can combine them in different configurations:

(a) To obtain a total resistance of 4 Ω: Connect the 2 Ω resistor and 3 Ω resistor in parallel, then connect this parallel combination in series with the 6 Ω resistor.

       ----[2Ω]----[3Ω]----
      |                     |
      -----------------------[6Ω]----  Total Resistance = 4Ω

(b) To obtain a total resistance of 1 Ω: Connect the 2 Ω resistor and 3 Ω resistor in series, then connect this series combination in parallel with the 6 Ω resistor.

       ----[2Ω]----[3Ω]----
      |                     |
      -----------------------[6Ω]----  Total Resistance = 1Ω

These configurations ensure that the overall resistance matches the desired values of 4 Ω and 1 Ω, respectively.

49.

To determine the highest and lowest resistance that can be obtained by combining four resistors with resistances of 4 Ω, 8 Ω, 12 Ω, and 24 Ω, we need to consider their combinations in series and parallel configurations.

(a) To find the highest resistance, we would connect all the resistors in series because the total resistance in a series connection is the sum of the individual resistances: Total resistance = $4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω$

So, the highest resistance obtained by combining these four resistors is $48 Ω$ .

(b) To find the lowest resistance, we would connect all the resistors in parallel because the total resistance in a parallel connection is given by the reciprocal of the sum of the reciprocals of the individual resistances: $\frac{1}{�_{total}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{24}$

Now, we calculate the total resistance: $\frac{1}{�_{total}} = \frac{6 + 3 + 2 + 1}{24} = \frac{12}{24} = \frac{1}{2}$

$�_{total} = 2 Ω$

So, the lowest resistance obtained by combining these four resistors is $2 Ω$ .

50.

51.To find the smallest and largest resistance that can be made using one hundred 1 Ω resistors, we need to consider their combinations in series and parallel configurations.

Smallest Resistance: To obtain the smallest resistance, we would connect all the resistors in parallel because the total resistance in a parallel connection is given by the reciprocal of the sum of the reciprocals of the individual resistances: $\frac{1}{�_{total}} = 100 \times \frac{1}{1} = 100$

$�_{total} = \frac{1}{100} Ω$

So, the smallest resistance obtained by combining these one hundred 1 Ω resistors is $\frac{1}{100} Ω$ .

Largest Resistance: To find the largest resistance, we would connect all the resistors in series because the total resistance in a series connection is the sum of the individual resistances: $�_{total} = 100 \times 1 = 100 Ω$

So, the largest resistance obtained by combining these one hundred 1 Ω resistors is $100 Ω$ .

52.To make a 250 Ω resistor using 100 Ω resistors, we need to combine them in a configuration that results in the desired resistance.

One way to achieve this is by connecting resistors in series and parallel combinations. Here's how we can do it:

Connect two 100 Ω resistors in series. This would give us a total resistance of 200 Ω.
Then, connect these two 100 Ω resistors in parallel with a single 100 Ω resistor.

This arrangement would look like this:

       ----[100Ω]----[100Ω]----
      |                          |
      --------------------------[100Ω]---- Total Resistance = 250Ω

In this configuration, the two 100 Ω resistors in series add up to 200 Ω, and when combined in parallel with another 100 Ω resistor, it results in a total resistance of 250 Ω

53.

54. First, let's calculate the equivalent resistance for the four resistances of 16 ohms each when connected in parallel.

When resistors are connected in parallel, the equivalent resistance ( $�_{parallel}$ ) can be calculated using the formula:

$\frac{1}{�_{parallel}} = \frac{1}{�_{1}} + \frac{1}{�_{2}} + \frac{1}{�_{3}} + \frac{1}{�_{4}}$

Given that all resistances are 16 ohms, we have:

$\frac{1}{�_{parallel}} = \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16}$ $\frac{1}{�_{parallel}} = 4 \times \frac{1}{16}$ $\frac{1}{�_{parallel}} = \frac{4}{16}$ $\frac{1}{�_{parallel}} = \frac{1}{4}$

$�_{parallel} = 4 Ω$

So, when four 16 ohm resistances are connected in parallel, the equivalent resistance is $4 Ω$ .

Now, these four equivalent resistances of $4 Ω$ each are connected in series. In a series connection, the total resistance is the sum of the individual resistances.

$�_{series} = 4 Ω + 4 Ω + 4 Ω + 4 Ω$ $�_{series} = 16 Ω$

Therefore, the total resistance when four resistances of 16 ohms each are connected in parallel, and four such combinations are connected in series, is $16 Ω$ .

55.

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Class 10 Chapter 1 Electricity Physics Solution S.Chand. Part 5

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