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Chapter 28: Problem 22

Three resistors, of \(8.0 \Omega, 12 \Omega\), and \(24 \Omega\), are in parallel, and a current of \(20 \mathrm{~A}\) is drawn by the combination. Determine \((a)\) the potential difference across the combination and \((b)\) the current through each resistance.

Short Answer

Expert verified
The potential difference is 80 V; currents are 10 A, 6.67 A, 3.33 A.

Step by step solution

01

Understand Parallel Resistor Formula

In a parallel circuit, the total or equivalent resistance \( R_{eq} \) is found using the formula: \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \), where \( R_1 = 8.0 \, \Omega \), \( R_2 = 12 \, \Omega \), and \( R_3 = 24 \, \Omega \).
02

Calculate Equivalent Resistance

Apply the formula for \( R_{eq} \) in parallel circuits:\[ \frac{1}{R_{eq}} = \frac{1}{8} + \frac{1}{12} + \frac{1}{24} \]Calculate each fraction:\[ \frac{1}{R_{eq}} = \frac{3}{24} + \frac{2}{24} + \frac{1}{24} \]Add the fractions:\[ \frac{1}{R_{eq}} = \frac{6}{24} = \frac{1}{4} \]Thus, \( R_{eq} = 4 \, \Omega \).
03

Calculate Potential Difference Across Combination

Use Ohm's Law, \( V = I \cdot R_{eq} \). Given that the total current \( I = 20 \, A \), and \( R_{eq} = 4 \, \Omega \):\[ V = 20 \times 4 = 80 \, V \].The potential difference across the combination is \( 80 \, V \).
04

Calculate Current Through Each Resistor

Use Ohm's Law for each resistor to find the current passing through it. For each resistor \( V = I_n \cdot R_n \):- For the 8 \( \Omega \) resistor, \( I_1 = \frac{V}{R_1} = \frac{80}{8} = 10 \, A \).- For the 12 \( \Omega \) resistor, \( I_2 = \frac{V}{R_2} = \frac{80}{12} \approx 6.67 \, A \).- For the 24 \( \Omega \) resistor, \( I_3 = \frac{V}{R_3} = \frac{80}{24} \approx 3.33 \, A \).

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Key Concepts

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

Ohm's Law
Ohm鈥檚 Law is a fundamental principle used extensively in electrical engineering and physics. It states that the voltage (\( V \)) across a resistor is directly proportional to the current (\( I \)) flowing through it and the resistance (\( R \)) of the resistor itself. This is succinctly expressed in the formula:
  • \( V = I \cdot R \)
In the context of parallel circuits, Ohm鈥檚 Law helps determine the potential difference across the whole circuit and across each individual resistor.

In the exercise provided, given a total current of 20 A and an equivalent resistance of 4 惟 for the parallel circuit, Ohm鈥檚 Law lets us find the voltage across the combination:
  • \( V = 20 \cdot 4 = 80 \text{ V} \)
This means each resistor in the parallel circuit experiences the same potential difference of 80 V.
Equivalent Resistance
Equivalent resistance in a parallel circuit refers to the resistance that could replace the actual resistors without changing the circuit's overall properties. This can be counter-intuitive since adding more resistors in parallel actually reduces the total resistance.

The formula for finding equivalent resistance \( R_{eq} \) in parallel circuits is:
  • \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]
In our exercise, the resistors are 8 惟, 12 惟, and 24 惟. Plugging these into the formula:
  • \[ \frac{1}{R_{eq}} = \frac{1}{8} + \frac{1}{12} + \frac{1}{24} \]
  • This simplifies to \[ \frac{1}{R_{eq}} = \frac{6}{24} = \frac{1}{4} \]
  • Thus, \( R_{eq} = 4 \, \Omega \)
This means a single 4 惟 resistor could replace the three parallel resistors without affecting the total current drawn.
Current Division in Parallel Circuits
When resistors are connected in parallel, the total current in the circuit divides between the branches. Each branch gets a portion of the total current, and the larger the resistance, the smaller the current it receives. This phenomenon is called current division.

To find the current through a specific resistor in parallel, we rearrange Ohm鈥檚 Law:
  • \( I_n = \frac{V}{R_n} \)
In the given exercise, the total voltage across each resistor is 80 V. Here's how the current divides among the resistors:
  • For the 8 惟 resistor, \( I_1 = \frac{80}{8} = 10 \, \text{A} \)
  • For the 12 惟 resistor, \( I_2 = \frac{80}{12} \approx 6.67 \, \text{A} \)
  • For the 24 惟 resistor, \( I_3 = \frac{80}{24} \approx 3.33 \, \text{A} \)
The sum of these currents should equal the total current, validating the distribution:
  • \( 10 + 6.67 + 3.33 = 20 \, \text{A} \)

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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 \(120-\mathrm{V}\) house circuit has the following light bulbs turned on: \(40.0 \mathrm{~W}, 60.0 \mathrm{~W}\), and \(175.0 \mathrm{~W}\). Find the equivalent resistance of these lights. House circuits are so constructed that each device is connected in parallel with the others. From \(\mathrm{P}=V I=V^{2} / R\), for the first bulb $$ R_{1}=\frac{V^{2}}{P_{1}}=\frac{(120 \mathrm{~V})^{2}}{40.0 \mathrm{~W}}=360 \Omega $$ Similarly, \(R_{2}=240 \Omega\) and \(R_{3}=192 \Omega\). Because devices in a house circuit are in parallel, $$ \frac{1}{R_{\mathrm{eq}}}=\frac{1}{360 \Omega}+\frac{1}{240 \Omega}+\frac{1}{192 \Omega} \quad \text { or } \quad R_{\mathrm{eq}}=82.3 \Omega $$ As a check, note that the total power drawn from the line is \(40.0 \mathrm{~W}+60.0 \mathrm{~W}+75.0 \mathrm{~W}=75.0 \mathrm{~W}\). Then, using \(\mathrm{P}=V^{2} / R\) $$ R_{\mathrm{eq}}=\frac{V^{2}}{\text { total power }}=\frac{(120 \mathrm{~V})^{2}}{175.0 \mathrm{~W}}=82.3 \Omega $$

Find the potential difference between points- \(P\) and \(-Q\) in Fig. \(28-8\). Which point is at the higher potential? From the result of Problem \(28.11\), the currents through \(P\) and \(Q\) are $$ \begin{array}{l} I_{P}=\frac{2 \Omega+18 \Omega}{10 \Omega+5 \Omega+2 \Omega+18 \Omega}(7.0 \mathrm{~A})=4.0 \mathrm{~A} \\ I_{Q}=\frac{10 \Omega+5 \Omega}{10 \Omega+5 \Omega+2 \Omega+18 \Omega}(7.0 \mathrm{~A})=3.0 \mathrm{~A} \end{array} $$ Now we start at point- \(P\) and go through point- \(a\) to point- \(Q\), to find Voltage change from \(P\) to \(Q=+(4.0 \mathrm{~A})(10 \Omega)-(3.0 \mathrm{~A})(2 \Omega)=+34 \mathrm{~V}\) (Notice that we go through a potential rise from \(P\) to \(a\) because we are going against the current. From \(a\) to \(Q\) there is a drop.) Therefore, the voltage difference between \(P\) and \(Q\) is \(34 \mathrm{~V}\), with \(Q\) being at the higher potential.

Two resistors, of \(4.00 \Omega\) and \(12.0 \Omega\), are connected in parallel across a 22-V battery having internal resistance \(1.00 \Omega\). Compute \((a)\) the battery current, \((b)\) the current in the \(4.00-\Omega\) resistor, \((c)\) the terminal voltage of the battery, \((d)\) the current in the \(12.0-\Omega\) resistor.

A voltmeter is to deflect full scale for a potential difference of \(5.000 \mathrm{~V}\) across it and is to be made by connecting a resistor \(R_{x}\) in series with a galvanometer. The \(80.00-\Omega\) galvanometer deflects full scale for a potential of \(20.00 \mathrm{mV}\) across it. Find \(R_{x}\). When the galvanometer is deflecting full scale, the current through it is $$ I=\frac{V}{R}=\frac{20.00 \times 10^{-3} \mathrm{~V}}{80.00 \Omega}=2.500 \times 10^{-4} \mathrm{~A} $$ When \(R_{x}\) is connected in series with the galvanometer, we wish \(I\) to be \(2.500 \times 10^{-4} \mathrm{~A}\) for a potential difference of \(5.000 \mathrm{~V}\) across the combination. Hence, \(V=I R\) becomes $$ 5.000 \mathrm{~V}=\left(2.500 \times 10^{-4} \mathrm{~A}\right)\left(80.00 \Omega+R_{x}\right) $$ from which \(R_{x}=19.92 \mathrm{k} \Omega\).
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