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Chapter 28: Problem 21

How many \(160-\Omega\) resistors (in parallel) are required to carry a total of \(5.0 \mathrm{~A}\) on a \(100-\mathrm{V}\) line?

Short Answer

Expert verified
8 resistors are needed.

Step by step solution

01

Understanding the Problem

To solve this problem, we need to determine how many resistors in parallel will be necessary to achieve a total current of 5.0 A across a 100 V supply with each resistor having a resistance of 160 惟.
02

Calculate Total Resistance Required

Using Ohm鈥檚 law, the total resistance required for the circuit is calculated by: \[ V = IR \] Rearranging gives: \[ R = \frac{V}{I} = \frac{100}{5.0} = 20 \Omega \] So, the entire parallel circuit needs to effectively have a resistance of 20 惟.
03

Formula for Parallel Resistance

For resistors in parallel, the formula is: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \] If each resistor is 160 惟 and they are all identical, the formula simplifies to: \[ \frac{1}{R_{eq}} = n \cdot \frac{1}{160} \] where n is the number of resistors.
04

Determine Number of Resistors

Given: \[ R_{eq} = 20 \Omega \] Thus, \[ \frac{1}{20} = n \cdot \frac{1}{160} \] Solving for n gives: \[ n = \frac{160}{20} = 8 \] We need 8 resistors of 160 惟 each in parallel to achieve the desired total resistance.

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Key Concepts

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

Ohm's law
Ohm's Law is fundamental in understanding how voltage, current, and resistance interrelate in electric circuits. The law is expressed by the equation: \[ V = I \cdot R \]where:
  • \( V \) is the voltage across the conductor measured in volts (V),
  • \( I \) is the current flowing through the conductor in amperes (A),
  • \( R \) is the resistance of the conductor measured in ohms (惟).
This equation tells us that if two of the quantities are known, the third can be calculated. For instance, if you know the voltage and the current, you can find the resistance by rearranging the formula to: \[ R = \frac{V}{I} \]In the problem, using Ohm鈥檚 Law, we calculated the total resistance needed to ensure a 5.0 A current under a 100 V supply to be 20 惟. This teaches us how resistance, current, and voltage are directly related to one another.
resistor network
A resistor network is a combination of multiple resistors. These can be arranged in various configurations like series or parallel, each having different rules on how they combine to affect the total current, voltage, and resistance in a circuit. In the given exercise, we dealt with a parallel resistor network. In a parallel configuration:
  • The voltage across each resistor is the same.
  • The total current is the sum of the individual currents through each resistor.
  • The reciprocal of the total resistance ( \( R_{eq} \)) is the sum of the reciprocals of each individual resistance ( \( R_1, R_2, ..., R_n \)).
This concept is particularly useful because putting resistors in parallel reduces the overall resistance. In our exercise, we used this principle to arrange 160 惟 resistors in parallel until we reached an equivalent resistance of 20 惟, thus allowing for the desired amount of current to flow.
resistance calculation
Resistance calculation in parallel circuits often requires the use of reciprocal mathematics. The formula for calculating the equivalent resistance in a parallel resistor network is:\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} \]This might look complex at first, but it becomes straightforward once you're familiar with reciprocals. In our problem:
  • We needed 160 惟 resistors in parallel to achieve a total equivalent resistance of 20 惟.
  • Using the formula, we found the number \( n \) of identical resistors required by setting \( n \cdot \frac{1}{160} = \frac{1}{20} \).
  • By solving, we determined that 8 resistors were needed.
This process shows that by understanding the rule of reciprocal calculation in parallel networks, one can easily determine how many resistors to use to achieve a specific resistance and current.

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Most popular questions from this chapter

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}$

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 \(R_{x^{\circ}}\) 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 \(R_{x}=19.92 \mathrm{k} \Omega\).

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 \(4.0 \Omega\) and \(8.0 \Omega(a)\) in series and \((b)\) in parallel.

Show that if two resistors are connected in parallel, the rates at which they produce thermal energy vary inversely as their resistances.
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