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Chapter 19: Problem 7

(II) How many 10-\(\Omega\) resistors must be connected in series to give an equivalent resistance to five 100-\(\Omega\) resistors connected in parallel?

Short Answer

Expert verified
You need 2 resistors in series.

Step by step solution

01

Calculate the equivalent resistance of 100-惟 resistors in parallel

When resistors are connected in parallel, the formula for equivalent resistance \( R_p \) is \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \). Since all resistors have the same resistance (100 惟), the formula becomes \( \frac{1}{R_p} = \frac{n}{R} \) where \( n \) is the number of resistors, and \( R \) is the resistance of each resistor. Here, \( n = 5 \) and \( R = 100 \, \Omega \), so \( \frac{1}{R_p} = \frac{5}{100} = \frac{1}{20} \). Thus, \( R_p = 20 \, \Omega \).
02

Calculate the number of 10-惟 resistors needed in series

To achieve an equivalent resistance of \( 20 \, \Omega \) using resistors of \( 10 \, \Omega \) each, and knowing the total resistance \( R_s \) for resistors in series is calculated as \( R_s = n \times R \) where \( R \) is the resistance of each resistor. Set \( R_s = 20 \) and \( R = 10 \), then \( 20 = n \times 10 \). Solving this gives \( n = 2 \).
03

Final Conclusion

Two 10-惟 resistors connected in series will provide an equivalent resistance of 20 惟, which matches the resistance of five 100-惟 resistors when connected in parallel.

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Key Concepts

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

Resistance Calculation
When working with an electrical circuit, determining the total resistance is crucial. Calculating resistance depends on how resistors are arranged in the circuit. Two common configurations are series and parallel.
In series, resistors add up directly to provide a total resistance. When in parallel, the reciprocal of their resistance values adds up. Understanding these arrangements helps solve problems related to combining resistors to achieve a desired resistance.
Resistors act like obstacles for electric current. The total resistance affects how much current flows through the circuit for a given voltage. Thus, accurate resistance calculation is vital for ensuring circuits operate efficiently and safely. By combining resistors in different ways, we can create circuits with precise resistance values.
Resistors in Series
Resistors in series are straightforward to handle because their resistances add up. Picture a single, longer resistor formed by all the smaller resistors placed in a row. The total series resistance, denoted by \( R_s \), is simply the sum of the individual resistances:

\[ R_s = R_1 + R_2 + R_3 + \ldots + R_n \]
This means if you have several 10-惟 resistors in series, their total resistance is the number of resistors multiplied by 10. For example, connecting two 10-惟 resistors in series provides a resistance of 20 惟.
This configuration is useful when a higher resistance value is needed, and only lower resistance resistors are available. Increasing total resistance reduces the current flowing through the circuit, useful for protecting components from excessive current.
Resistors in Parallel
Resistors in parallel act differently. They help decrease the total resistance. When connected this way, the formula for finding the equivalent resistance \( R_p \) is:

\[ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \]
This can be simplified as \( \frac{n}{R} \) if all resistors have the same value, where \( n \) is the number of resistors. Connecting multiple resistors in parallel offers paths for the current, effectively reducing total resistance. For instance, five 100-惟 resistors in parallel provide 20 惟.
This setup is excellent when a lower resistance is required but only higher resistance resistors are present. Lowering resistance increases current flow, useful in circuits requiring more power, like heating devices or powerful LEDs.

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Most popular questions from this chapter

What are the values of effective capacitance which can be obtained by connecting four identical capacitors, each having a capacitance \(C\)?

Suppose that a person's body resistance is 950 \(\Omega\) (moist skin). (\(a\)) What current passes through the body when the person accidentally is connected to 120 V? (\(b\)) If there is an alternative path to ground whose resistance is 25 \(\Omega\), what then is the current through the body? (\(c\)) If the voltage source can produce at most 1.5 A, how much current passes through the person in case (\(b\))?

(II) Three 1.70-k\(\Omega\) resistors can be connected together in four different ways, making combinations of series and/or parallel circuits. What are these four ways, and what is the net resistance in each case?

How many \(\frac{1}{2}\)-W resistors, each of the same resistance, must be used to produce an equivalent 3.2-k\(\Omega\), 3.5-W resistor? What is the resistance of each, and how must they be connected? Do not exceed \(P = \frac{1}{2}\)W in each resistor.

A \(\textbf{Wheatstone bridge}\) is a type of "bridge circuit" used to make measurements of resistance. The unknown resistance to be measured, \(R_x,\) is placed in the circuit with accurately known resistances \(R_1, R_2,\) and \(R_3\) (Fig. 19-76). One of these, is a variable resistor which is adjusted so that when the switch is closed momentarily, the ammeter A shows zero current flow. The bridge is then said to be balanced. (\(a\)) Determine \(R_x\) in terms of \(R_1, R_2,\) and \(R_3\). (\(b\)) If a Wheatstone bridge is "balanced" when \(R_1 =\) 590 \(\Omega\), \(R_2 =\) 972 \(\Omega\), and \(R_3 =\) 78.6 \(\Omega\), what is the value of the unknown resistance?
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