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Derive the formula for the equivalent resistance \(R_{\mathrm{eq}}\) of resistors \(R_{1}\), \(R_{2}\), and \(R_{3}(a)\) in series and \((b)\) in parallel, as shown in \(\underline{\text \)\underline{1}(a)\( and \)(b)\(. (a) For the series network, $$ V_{a d}=V_{a b}+V_{b c}+V_{c d}=I R_{1}+I R_{2}+I R_{3} $$ since the current \)I\( is the same in all three resistors. Dividing by \)I\( gives $$ \frac{V_{a d}}{I}=R_{1}+R_{2}+R_{3} \quad \text { or } \quad R_{\mathrm{eq}}=R_{1}+R_{2}+R_{3} $$ since \)V_{a d} / I\( is by definition the equivalent resistance \)R_{\mathrm{eq}}\( of the network. (b) The p.d. is the same for all three resistors, whence $$ I_{1}=\frac{V_{a b}}{R_{1}} \quad I_{2}=\frac{V_{a b}}{R_{2}} \quad I_{3}=\frac{V_{a b}}{R_{3}} $$ Since the line current \)I\( is the sum of the branch currents, $$ I=I_{1}+I_{2}+I_{3}=\frac{V_{a b}}{R_{1}}+\frac{V_{d b}}{R_{2}}+\frac{V_{a b}}{R_{3}} $$ Dividing by \)V_{a b}\( gives $$ \frac{I}{V_{a b}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}} \quad \text { or } \quad \frac{1}{R_{\mathrm{eq}}}=\frac{1}{R_{\mathrm{1}}}+\frac{1}{R_{2}}+\frac{1}{R_{3}} $$ since \)V_{a b} / I\( is by definition the equivalent resistance \)R_{\mathrm{eq}}$ of the network.

Short Answer

Expert verified
Series: \(R_{\mathrm{eq}} = R_1 + R_2 + R_3\); Parallel: \(\frac{1}{R_{\mathrm{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\).

Step by step solution

01

Understand Series Circuit Parameters

In a series circuit, resistors are connected end-to-end, and the total voltage across them is the sum of the individual voltages. The current, \(I\), is the same through each resistor, and this is used to find the equivalent resistance \(R_{\mathrm{eq}}\).
02

Apply Ohm’s Law to Series Circuit

Using Ohm's Law for each resistor, we can express the voltage across them as follows: \( V_{ad} = IR_1 + IR_2 + IR_3 \). Since \(I\) is constant, dividing both sides by \(I\) gives: \( \frac{V_{ad}}{I} = R_1 + R_2 + R_3 \). The equivalent resistance \(R_{\mathrm{eq}}\) is \( R_1 + R_2 + R_3 \).
03

Understand Parallel Circuit Parameters

In a parallel circuit, all resistors share the same voltage. The total current is the sum of currents through each resistor. Thus, the equivalent resistance is derived based on these currents.
04

Express Current in Parallel Circuit Using Ohm's Law

Each resistor in parallel has the same potential difference, \(V_{ab}\). Thus, currents through each resistor are \( I_1 = \frac{V_{ab}}{R_1} \), \( I_2 = \frac{V_{ab}}{R_2} \), and \( I_3 = \frac{V_{ab}}{R_3} \).
05

Sum Currents to Find Equivalent Resistance in Parallel

The total current \(I\) in the circuit is the sum of the branch currents: \( I = I_1 + I_2 + I_3 = \frac{V_{ab}}{R_1} + \frac{V_{ab}}{R_2} + \frac{V_{ab}}{R_3} \). Dividing by \(V_{ab}\), we get \( \frac{I}{V_{ab}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \).
06

Solve for Parallel Equivalent Resistance

Inverting both sides gives the formula for the equivalent resistance: \( \frac{1}{R_{\mathrm{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \). This defines the equivalent resistance of resistors in parallel.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Circuit
In a series circuit, resistors are connected in a single path. This means the current has only one path to take, flowing through each resistor one after the other.
A key characteristic of series circuits is that the current (\(I\)) remains constant throughout the entire circuit. As the current flows through each resistor, it experiences a voltage drop, resulting in the total voltage across the circuit equaling the sum of all individual voltage drops.
The equivalent resistance (\(R_{\mathrm{eq}}\)) for a series circuit can be calculated by simply adding up the resistance values of all connected resistors: \(R_{\mathrm{eq}} = R_1 + R_2 + R_3\). This formula shows how resistances in series add together to form a larger total resistance.
Parallel Circuit
Parallel circuits are a different beast altogether. Here, resistors are connected across the same two points, meaning each resistor has its own branch.
The advantage of parallel configurations is that they maintain the same potential difference or voltage across each resistor. However, the current divides, with a portion passing through each branch. The total current (\(I\)) is the sum of the branch currents.
To find the equivalent resistance of resistors in parallel, you apply the formula: \(\frac{1}{R_{\mathrm{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\). In this scenario, the equivalent resistance will always be less than the smallest resistor in the network. This opposite behavior to series circuits provides flexibility in how circuits can be utilized.
Ohm's Law
Ohm's Law is a fundamental principle essential to understanding electric circuits. It establishes the relationship between voltage (\(V\)), current (\(I\)), and resistance (\(R\)).
  • Voltage (\(V\)) is the electrical potential difference across a component.
  • Current (\(I\)) is the flow of electric charge through a conductor.
  • Resistance (\(R\)) is a measure of how much a component resists this flow.
Ohm's Law is expressed as \(V = IR\), where the voltage is the product of current and resistance. This relation is invaluable in both series and parallel circuits, aiding in determining how current and voltage distribute across resistors. Using Ohm's Law, you can calculate unknown variables, making it an essential tool for analyzing circuits.
Resistor Networks
Resistor networks are combinations of multiple resistors configured in series, parallel, or complex interconnections of both. Each configuration impacts the overall equivalent resistance and how current and voltage are managed.
By mastering how series and parallel circuits function, you can analyze and simplify complex resistor networks to determine the behavior of electrical systems.
  • Simplification: Break down complex networks into smaller, manageable series or parallel sections.
  • Calculation: Use the respective formulas for series (\(R_{\mathrm{eq}}=R_1+R_2+R_3\)) and parallel circuits (\(\frac{1}{R_{\mathrm{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\)) to find the equivalent resistance.
  • Analysis: Determine how changes in individual resistor values affect the overall circuit.
Acquiring these skills enables the design and troubleshooting of more intricate circuits, forming the backbone of practical electronics.

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Most popular questions from this chapter

What resistance must be placed in parallel with \(20 \Omega\) to make the combined resistance \(15 \Omega ?\)

A relay having a resistance of \(6.0 \Omega\) operates with a minimum current of \(0.030 \mathrm{~A}\). It is required that the relay operate when the current in the line reaches \(0.240 \mathrm{~A}\). What resistance should be used to shunt the relay?

Compute the equivalent resistance of \(4.0 \Omega\) and \(8.0 \Omega(a)\) in series and \((b)\) in parallel.

As shown in \(\underline a battery (internal resistance \)1 \Omega\( ) is connected in series with two resistors. Compute ( \)a\( ) the current in the circuit, (b) the p.d. across each resistor, and \)(c)\( the terminal p.d. of the battery. The circuit is redrawn so as to show the battery resistance. The resistors are in series, $$ R_{\mathrm{eg}}=5 \Omega+12 \Omega+1 \Omega=18 \Omega $$ Hence, the circuit is equivalent to the one shown Applying \)V=\mathrm{IR}\( \)(a)\( $$ I=\frac{V}{R}=\frac{18 \mathrm{~V}}{18 \Omega}=1.0 \mathrm{~A} $$ (b) Since \)I=1.0 \mathrm{~A}\(, we can find the p.d. from point- \)b\( to point- \)c\( as $$ V_{b c}=I R_{b c}=(1.0 \mathrm{~A})(12 \Omega)=12 \mathrm{~V} $$ and that from \)c\( to \)d\( as $$ V_{c d}=I R_{c d}=(1.0 \mathrm{~A})(5 \Omega)=5 \mathrm{~V} $$ Notice that \)I\( is the same at all points in a series circuit. (c) The terminal p.d. of the battery is the p.d. from \)a\( to \)e\(. Therefore, $$ \text { Terminal p.d. }=V_{b c}+V_{c d}=12+5=17 \mathrm{~V} $$ Or, we could start at \)e\( and keep track of the voltage changes as we go through the battery from \)e\( to \)a\(. Taking voltage drops as negative, Terminal p.d. \)=-I r+\varepsilon=-(1.0 \mathrm{~A})(1 \Omega)+18 \mathrm{~V}=17 \mathrm{~V}$

Three resistors, of \(40 \Omega, 60 \Omega\), and \(120 \Omega\), are connected in parallel, and this parallel group is connected in series with \(15 \Omega\) in series with \(25 \Omega\). The whole system is then connected to a 120 \(\mathrm{V}\) source. Determine \((a)\) the current in the \(25 \Omega,(b)\) the potential drop across the parallel group, (c) the potential drop across the 25 \(\Omega,(d)\) the current in the \(60 \Omega,(e)\) the current in the \(40 \Omega\).

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