11.6-Resistance of a System of Resistors

11.6-Resistance of a System of Resistors Important Formulae

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11.6-Resistance of a System of Resistors

The total resistance of a system of resistors depends on how the resistors are connected in the circuit. There are two main ways to connect resistors: series connection and parallel connection. Let’s explore both types of connections in detail.

Resistors in Series

When resistors are connected end to end, they are said to be connected in series. In a series connection, the current flowing through each resistor is the same. The total resistance $R_{\text{total}}$ is the sum of the individual resistances of the resistors:

Formula: $R_{\text{total}} = R_1 + R_2 + R_3 + \cdots + R_n$

In this case, the total resistance increases as more resistors are added in series. The reason is that each resistor restricts the flow of current, thus increasing the overall opposition to the current flow. The voltage across each resistor can be different depending on its resistance value, but the current through each remains the same.

Resistors in Parallel

When resistors are connected across the same two points, they are said to be connected in parallel. In a parallel connection, the voltage across each resistor is the same, but the total current divides among the resistors. The total resistance $R_{\text{total}}$ of resistors in parallel can be calculated using the reciprocal rule:

Formula: $\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots + \frac{1}{R_n}$

As more resistors are added in parallel, the total resistance decreases. This is because adding more pathways for the current to flow allows more current to pass through the circuit. The current through each resistor may differ, depending on their individual resistance, but the voltage remains the same across all resistors.

Combination of Series and Parallel Resistors

In most circuits, resistors are not purely in series or parallel. They may be arranged in a combination of series and parallel. In such cases, we need to calculate the total resistance step by step, starting by simplifying the parallel groups and then adding the resistances in series, or vice versa.

For example, if you have a combination of resistors where some are in series and others in parallel, first calculate the equivalent resistance of the parallel group, then add that to the series resistors to find the total resistance.

Example of a Combination Circuit

Consider three resistors: $R_1 = 4 \, \Omega$, $R_2 = 6 \, \Omega$, and $R_3 = 12 \, \Omega$. Suppose $R_1$ and $R_2$ are in parallel, and their combination is in series with $R_3$. The total resistance can be calculated as follows:

Step 1: Calculate the parallel resistance of $R_1$ and $R_2$:

$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12}$

Therefore, $R_{\text{parallel}} = \frac{12}{5} = 2.4 \, \Omega$.

Step 2: Add the parallel resistance to $R_3$ (in series):

$R_{\text{total}} = R_{\text{parallel}} + R_3 = 2.4 + 12 = 14.4 \, \Omega$.

Important Points to Remember
  • In series connection, the total resistance is the sum of the individual resistances.
  • In parallel connection, the total resistance is always less than the smallest individual resistance.
  • When resistors are connected in a combination of series and parallel, reduce the circuit step by step to find the total resistance.
  • The unit of resistance is the ohm ($\Omega$).

11.6-प्रतिरोध का एकत्रीकरण (Resistance of a System of Resistors)

जब कई प्रतिरोधक (resistors) एक साथ जोड़े जाते हैं, तो उनका कुल प्रतिरोध (equivalent resistance) निर्धारित किया जाता है। प्रतिरोधकों को जोड़ने के दो मुख्य तरीके होते हैं: अनुक्रम में (series) और समानांतर में (parallel)।

1. प्रतिरोधकों का अनुक्रम में जोड़ना (Resistors in Series)

जब प्रतिरोधक एक-दूसरे के साथ जुड़ते हैं, तो उन्हें अनुक्रम में जोड़ने पर कुल प्रतिरोध बढ़ता है। इसका कारण यह है कि धारा सभी प्रतिरोधकों से समान रूप से गुजरती है, और प्रत्येक प्रतिरोधक धारा को रोकता है।

यदि एक प्रणाली में $n$ प्रतिरोधक $R_1, R_2, R_3, \dots, R_n$ हैं, तो उनका कुल प्रतिरोध $R_{total}$ इस प्रकार होता है:

कुल प्रतिरोध (Total Resistance) :

$$R_{total} = R_1 + R_2 + R_3 + \dots + R_n$$

यहाँ, $R_1, R_2, R_3, \dots, R_n$ प्रतिरोधकों के व्यक्तिगत प्रतिरोध हैं।

2. प्रतिरोधकों का समानांतर में जोड़ना (Resistors in Parallel)

जब प्रतिरोधक समानांतर में जोड़े जाते हैं, तो कुल प्रतिरोध कम होता है। इस स्थिति में, धारा विभिन्न शाखाओं में विभाजित हो जाती है, और प्रत्येक शाखा का प्रतिरोध अलग होता है। समानांतर संयोजन में कुल प्रतिरोध घटता है क्योंकि धारा के लिए अधिक रास्ते उपलब्ध होते हैं।

यदि $n$ प्रतिरोधक $R_1, R_2, R_3, \dots, R_n$ समानांतर में जुड़ें, तो कुल प्रतिरोध $R_{total}$ का सूत्र है:

कुल प्रतिरोध (Total Resistance) :

$$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}$$

यहाँ, $R_1, R_2, R_3, \dots, R_n$ समानांतर में जुड़े प्रतिरोधकों के व्यक्तिगत प्रतिरोध हैं।

3. समानांतर और अनुक्रम में दोनों का संयोजन (Combination of Series and Parallel Resistors)

अक्सर प्रतिरोधक प्रणाली में दोनों प्रकार के संयोजन होते हैं, यानी कुछ प्रतिरोधक अनुक्रम में होते हैं और कुछ समानांतर में। ऐसे मामलों में, पहले अनुक्रम में जुड़े प्रतिरोधकों का कुल प्रतिरोध निकालें और फिर उसे समानांतर में जुड़े प्रतिरोधकों से जोड़ें, या इसके विपरीत।

इस प्रकार की समस्याओं को हल करने के लिए, चरण दर चरण प्रतिरोधकों का जोड़ने का तरीका अपनाना चाहिए। पहले अनुक्रम में जुड़े प्रतिरोधकों का कुल प्रतिरोध निकालें, फिर उसे समानांतर संयोजन में जोड़ें, और अंत में समग्र प्रतिरोध प्राप्त करें।

उदाहरण:

मान लीजिए तीन प्रतिरोधक $R_1 = 2 \, \Omega$, $R_2 = 3 \, \Omega$, और $R_3 = 6 \, \Omega$ दिए गए हैं। यदि $R_1$ और $R_2$ अनुक्रम में जुड़े हों और $R_3$ समानांतर में हो, तो पहले $R_1$ और $R_2$ का कुल प्रतिरोध निकालें:

$$R_{12} = R_1 + R_2 = 2 + 3 = 5 \, \Omega$$

अब, $R_{12}$ और $R_3$ का कुल प्रतिरोध निकालें (चूँकि ये समानांतर में हैं):

$$\frac{1}{R_{total}} = \frac{1}{R_{12}} + \frac{1}{R_3} = \frac{1}{5} + \frac{1}{6}$$

साधारण रूप से जोड़ने के बाद:

$$\frac{1}{R_{total}} = \frac{6 + 5}{30} = \frac{11}{30}$$

अतः,

$$R_{total} = \frac{30}{11} \approx 2.73 \, \Omega$$

इस प्रकार, कुल प्रतिरोध $2.73 \, \Omega$ है।

4. ओम का नियम (Ohm’s Law) और प्रतिरोध

ओम के नियम के अनुसार, किसी परिपथ में विद्युत धारा $I$ और वोल्टेज $V$ के बीच का संबंध निम्नलिखित है:

$$V = I \times R$$

जहाँ, $V$ वोल्टेज, $I$ विद्युत धारा, और $R$ प्रतिरोध है। यदि प्रतिरोध को जोड़ने की प्रक्रिया को सही से समझा जाए, तो ओम के नियम का उपयोग कर प्रणाली में धारा और वोल्टेज के संबंध को आसानी से हल किया जा सकता है।

Draw a schematic diagram of a circuit consisting of a battery of three ? cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and
a plug key, all connected in series.

Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Judge the equivalent resistance when the following are connected in parallel
(a)1 Ω and 10$^6$ Ω,
(b)1 Ω and 10$^3$ Ω, and 10$^6$ Ω.

Solution:

Equivalent Resistance in Parallel Connections

For resistors connected in parallel, the formula for equivalent resistance (Req) is given by:

1/Req = 1/R1 + 1/R2 + ...

(a) For 1 Ω and 106 Ω:

1/Req = 1/1 + 1/106 = 1 + 0.000001 = 1.000001

Req ≈ 1 Ω

(b) For 1 Ω, 103 Ω, and 106 Ω:

1/Req = 1/1 + 1/1000 + 1/106 = 1 + 0.001 + 0.000001 = 1.001001

Req ≈ 0.999 Ω

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An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water
filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source
that takes as much current as all three appliances, and what is the
current through it?

Solution:

Current and Resistance in Parallel Appliances

For appliances connected in parallel, the equivalent resistance (Req) is calculated as:

1/Req = 1/R1 + 1/R2 + 1/R3

For the electric lamp (100 Ω), toaster (50 Ω), and water filter (500 Ω):

1/Req = 1/100 + 1/50 + 1/500 = 0.01 + 0.02 + 0.002 = 0.032

Req = 1/0.032 = 31.25 Ω

The total current (I) through Req is:

I = V/Req = 220 V / 31.25 Ω ≈ 7.04 A

For the electric iron, R = V/I = 220 V / 7.04 A ≈ 31.25 Ω.

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What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Solution:

Advantages of Connecting Electrical Devices in Parallel

Connecting electrical devices in parallel with a battery offers several advantages. Firstly, each device receives the full voltage of the battery, ensuring optimal performance. Secondly, if one device fails, the others continue to function, enhancing reliability. Thirdly, devices can be turned on or off independently without affecting the entire circuit. Additionally, parallel connections allow for a greater overall current draw, accommodating devices with different power requirements. Lastly, this configuration prevents voltage drops across devices, maintaining consistent performance and efficiency across all connected devices.

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How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

Solution:

Connecting Resistors to Achieve Desired Resistance

To achieve a total resistance of 4 Ω, connect the 2 Ω and 6 Ω resistors in parallel. The formula for parallel resistance is 1/R_total = 1/R1 + 1/R2. Thus, 1/R_total = 1/2 + 1/6, which simplifies to R_total = 3 Ω. Adding the 3 Ω resistor in series gives a total of 4 Ω. For a total resistance of 1 Ω, connect the 2 Ω, 3 Ω, and 6 Ω resistors all in parallel. Applying the formula, 1/R_total = 1/2 + 1/3 + 1/6 results in R_total = 1 Ω.

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What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Solution:

Highest and Lowest Total Resistance of Coils

The highest total resistance from four coils of 4 Ω, 8 Ω, 12 Ω, and 24 Ω is achieved by connecting them in series. The total resistance is calculated by summing the individual resistances: 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω. The lowest total resistance is obtained by connecting all four coils in parallel. Using the formula 1/R_total = 1/R1 + 1/R2 + 1/R3 + 1/R4, we have 1/R_total = 1/4 + 1/8 + 1/12 + 1/24. Simplifying this gives R_total = 2 Ω.

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