/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Compute the equivalent resistanc... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
Series: 12.0 Ω Parallel: 2.67 Ω

Step by step solution

01

Understanding Series Resistance

When resistors are connected in series, the total or equivalent resistance is simply the sum of the individual resistances. This is because the current has only one path to take through each resistor.
02

Calculate Series Resistance

For resistors connected in series, the formula to calculate the equivalent resistance is: \[R_{series} = R_1 + R_2\]Substituting the given values, we have: \[R_{series} = 4.0 \, \Omega + 8.0 \, \Omega = 12.0 \, \Omega\]
03

Understanding Parallel Resistance

In a parallel circuit, the voltage across each resistor is the same, and the equivalent resistance is found using the reciprocal formula. This results in a lower equivalent resistance compared to any individual resistor in the circuit due to multiple current paths.
04

Calculate Parallel Resistance

The formula for equivalent resistance for resistors in parallel is:\[\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2}\]Substitute the given resistance values:\[\frac{1}{R_{parallel}} = \frac{1}{4.0 \, \Omega} + \frac{1}{8.0 \, \Omega}\]Simplifying further, we calculate:\[\frac{1}{R_{parallel}} = \frac{2}{8.0 \, \Omega} + \frac{1}{8.0 \, \Omega} = \frac{3}{8.0 \, \Omega}\]Thus, \[R_{parallel} = \frac{8.0}{3} \, \Omega \approx 2.67 \, \Omega\]

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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, every component is connected end-to-end, forming a single path for electrical current to flow. Imagine a chain of Christmas lights: if one bulb goes out, the circuit is broken, and all bulbs go dark. This is because the current has to pass through each resistance sequentially. In such systems, the current is consistent across each component, while the voltage across each resistor adds up to the total voltage supplied. To find the equivalent resistance in a series circuit, you simply add up the individual resistances.
  • The formula is: \[ R_{series} = R_1 + R_2 + ... + R_n \]
  • This was used to find the equivalent resistance of 4.0 \( \Omega \) and 8.0 \( \Omega \) resistors as \[ 12.0 \Omega \] in the exercise.
This simple addition is due to the fact that there is only one path for the electrons to travel, making it easy to understand and calculate the total resistance in a series circuit.
parallel circuit
In contrast to series circuits, parallel circuits allow current to flow through multiple paths. Think of a house's electrical wiring: different appliances can operate independently because they are on separate paths. In such circuits, all components share the same voltage, but the current can vary, dividing amongst the different paths. The equivalent resistance in a parallel circuit is always less than the smallest resistance involved, thanks to the way electric current is distributed.
  • The formula for calculating this is:\[ \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n} \]
  • For our exercise, this calculation led to an equivalent resistance of around \[ 2.67 \Omega \] which is lower than the smallest resistor in the problem, \( 4.0 \Omega \).
Parallel circuits ensure that even if one component fails, current can still flow through the other paths, maintaining the operation of other parts of the circuit.
equivalent resistance
Equivalent resistance is a concept used to simplify complex circuits by reducing a group of resistors into a single resistor that has the same effect on the overall circuit. Whether resistors are combined in series or parallel, the goal is to find out how they collectively affect the circuit's current and voltage. In simple terms, it is asking, "What is the single resistance that can replace these interconnected resistors without altering the circuit's behavior?"
  • For series circuits, add the resistances directly as they line up in sequence.
  • For parallel circuits, compute using the reciprocal formula to accommodate the multiple pathways available for current flow.
  • This helps in designing and analyzing circuits by simplifying them into more manageable parts.
Once you understand how to calculate equivalent resistance, even complex networks of resistors become much easier to troubleshoot or design.
Ohm's Law
Ohm's Law is a fundamental principle in the study of electricity, relating the voltage \( V \), current \( I \), and resistance \( R \) of a circuit. It provides a critical formula for understanding how these variables interdepend:\[ V = I \times R \]This equation tells us that the current flowing through a circuit between two points is directly proportional to the voltage across those two points and inversely proportional to the resistance.
  • In a series circuit, the same current flows through all components, but the voltage is divided among them.
  • In parallel circuits, the voltage remains constant across all branches, while the current can vary.
  • Utilizing Ohm's Law allows engineers and students to calculate one of the variables if the other two are known.
Ohm's Law is invaluable for solving any electrical problem as it links the major elements of electrical circuit behavior, offering clear insight into how circuit changes will affect overall performance.

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Most popular questions from this chapter

A current of \(5.0\) A flows into the circuit in point- \(a\) and out at point-b. (a) What is the potential difference from \(a\) to b? (b) How much current flows through the \(12.0-\Omega\) resistor? we found that the equivalent resistance for this combination is \(11.6 \Omega\), and we are told the current through it is 5.0 A. (a) Voltage drop from \(a\) to \(b=I R_{\mathrm{eq}}=(5.0 \mathrm{~A})(11.6 \Omega)=58 \mathrm{~V}\) (b) The voltage drop from \(a\) to \(c\) is \((5.0 \mathrm{~A})(9.0 \Omega) 45 \mathrm{~V}\). Hence, from part \((a)\), the voltage drop from \(c\) to \(b\) is $$ 58 \mathrm{~V}-45 \mathrm{~V}=13 \mathrm{~V} $$ and the current in the \(12.0-\Omega\) resistor is $$ I_{12}=\frac{V}{R}=\frac{13 \mathrm{~V}}{12 \Omega}=1.1 \mathrm{~A} $$

Compute the equivalent resistance of \((a) 3.0 \Omega, 6.0 \Omega\), and \(9.0 \Omega\) in parallel; (b) \(3.0 \Omega, 4.0 \Omega, 7.0 \Omega, 10.0 \Omega\), and \(12.0 \Omega\) in parallel; (c) three 33-\Omega heating elements in parallel; \((d\) ) twenty \(100-\Omega\) lamps in parallel.

For the circuit shown find the current in each resistor and the current drawn from the 40 - \(V\) source. Notice that the p.d. from \(a\) to \(b\) is \(40 \mathrm{~V}\). Therefore, the p.d. across each resistor is \(40 \mathrm{~V}\). Then, $$ I_{2}=\frac{40 \mathrm{~V}}{2.0 \Omega}=20 \mathrm{~A} \quad I_{5}=\frac{40 \mathrm{~V}}{5.0 \Omega}=8.0 \mathrm{~A} \quad I_{8}=\frac{40 \mathrm{~V}}{8.0 \Omega}=5.0 \mathrm{~A} $$ Because \(I\) splits into three currents: $$ I=I_{2}+I_{5}+I_{8}=20 \mathrm{~A}+8.0 \mathrm{~A}+5.0 \mathrm{~A}=33 \mathrm{~A} $$

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.

Find the equivalent resistance between points- \(a\) and \(-b\) for the combination shown The \(3.0-\Omega\) and \(2.0-\Omega\) resistors are in series and are equivalent to a 5.0-\Omega resistor. The equivalent \(5.0 \Omega\) is in parallel with the \(6.0 \Omega\), and their equivalent, \(R_{1}\), is $$ \frac{1}{R_{1}}=\frac{1}{5.0 \Omega}+\frac{1}{6.0 \Omega}=0.20+0.167=0.367 \Omega^{-1} \quad \text { or } \quad R_{1}=2.73 \Omega $$ The circuit thus far reduced is shown The \(7.0 \Omega\) and \(2.73 \Omega\) are equivalent to \(9.73 \Omega\). Now the \(5.0 \Omega\), \(12.0 \Omega\), and \(9.73 \Omega\) are in parallel, and their equivalent, \(R_{2}\), is $$ \frac{1}{R_{2}}=\frac{1}{5.0 \Omega}+\frac{1}{12.0 \Omega}+\frac{1}{9.73 \Omega}=0.386 \Omega^{-1} \quad \text { or } \quad R_{2}=2.6 \Omega $$ This \(2.6 \Omega\) is in series with the \(9.0-\Omega\) resistor. Therefore, the equivalent resistance of the combination is \(9.0 \Omega+2.6 \Omega=11.6\) \(\Omega\)

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