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Chapter 27: Problem 19

A total resistance of \(3.00 \Omega\) is to be produced by connecting an unknown resistance to a \(12.0 \Omega\) resistance. (a) What must be the value of the unknown resistance, and (b) should it be connected in series or in parallel?

Short Answer

Expert verified
The unknown resistance must be \( 4.00 \, \Omega \) and should be connected in parallel.

Step by step solution

01

Understanding the Total Resistance for Series Connection

For resistors in series, the total resistance is the sum of the individual resistances. If the total resistance is given as \( R_t = 3.00 \, \Omega \) and one resistor is \( 12.0 \, \Omega \), then for a series connection the formula is \( R_t = R_1 + R_2 \). Thus, \( 3.00 = 12.0 + R \). This is not possible since we can鈥檛 have a total resistance of \( 3.00 \, \Omega \) by adding \( 12.0 \, \Omega \). Thus, the resistors cannot be connected in series.
02

Understanding the Total Resistance for Parallel Connection

For resistors in parallel, the total resistance is calculated using \( \frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} \). Here, we know \( R_t = 3.00 \, \Omega \) and \( R_1 = 12.0 \, \Omega \). Substituting these values in: \( \frac{1}{3.00} = \frac{1}{12.0} + \frac{1}{R} \).
03

Solve for Unknown Resistance in Parallel

From the equation \( \frac{1}{3.00} = \frac{1}{12.0} + \frac{1}{R} \), solve for \( \frac{1}{R} \). First simplify: \( \frac{1}{R} = \frac{1}{3.00} - \frac{1}{12.0} \). This gives: \( \frac{1}{R} = \frac{4}{12} - \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \).
04

Calculate the Value of the Unknown Resistance

Now, solve for \( R \): \( \frac{1}{R} = \frac{1}{4} \) implies \( R = 4.00 \, \Omega \). So, the unknown resistance must be \( 4.00 \, \Omega \).

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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, resistors are connected end-to-end, forming a single path for current to flow. The total resistance in a series circuit is simply the sum of all individual resistances. This is because the current, which is constant throughout any series circuit, must pass through each resistor sequentially.
  • The formula for calculating total resistance: \[ R_t = R_1 + R_2 + ... + R_n \]where \( R_t \) is the total resistance, and \( R_1, R_2, \) ... \( R_n \) are the individual resistances.
  • This means that adding more resistors in series increases the total resistance.
In the problem you provided, a series connection isn't feasible because a 3.00 \( \Omega \) total resistance cannot be achieved by adding a resistor to the 12.0 \( \Omega \). The sum would always exceed 12.0 \( \Omega \), which is not what we need.
Parallel Circuit
A parallel circuit features components connected alongside each other. This means each pathway has its own direct connection to the voltage source. For parallel resistors, unlike a series circuit, the total resistance decreases as more resistors are added.
  • Total resistance for a parallel connection is less than the smallest resistance in the circuit.
  • The formula used is:\[ \frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n} \]
  • This formula shows that the reciprocals of each individual resistance add up to give the reciprocal of the total resistance.
In the exercise, the resistors must be connected in parallel to achieve a total resistance of 3.00 \( \Omega \). Solving with the given formula, the unknown resistor was calculated to be 4.00 \( \Omega \). This works because the reciprocal of the total resistance correctly fits the requirement for creating a 3.00 \( \Omega \) circuit.
Ohm's Law
Ohm's Law is a fundamental principle in electronics and physics. It helps us understand the relationship between voltage, current, and resistance in electrical circuits. This key law is expressed through the simple equation:\[ V = IR \]where:
  • \( V \) represents voltage across the resistance
  • \( I \) is the current flowing through the resistance
  • \( R \) is the resistance
This formula allows us to calculate one variable if the other two are known, making it extremely useful for solving many circuit problems. In our problem, Ohm's Law might not be directly needed since we're primarily solving for resistance values, but it remains an essential tool in understanding how resistors in different configurations affect total resistance and overall circuit behavior. By mastering this concept, you can effectively analyze and design electrical circuits.

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Most popular questions from this chapter

A wire of resistance \(5.0 \Omega\) is connected to a battery whose emf \(\mathscr{E}\) is \(2.0 \mathrm{~V}\) and whose internal resistance is \(1.0 \Omega\). In \(2.0 \mathrm{~min}\), how much energy is (a) transferred from chemical form in the battery, (b) dissipated as thermal energy in the wire, and (c) dissipated as thermal energy in the battery?

A \(15.0 \mathrm{k} \Omega\) resistor and a capacitor are connected in series, and then a \(12.0 \mathrm{~V}\) potential difference is suddenly applied across them. The potential difference across the capacitor rises to \(5.00 \mathrm{~V}\) in \(1.30 \mu \mathrm{s}\). (a) Calculate the time constant of the circuit. (b) Find the capacitance of the capacitor.

Four \(18.0 \Omega\) resistors are connected in parallel across a \(25.0 \mathrm{~V}\) ideal battery. What is the current through the battery?

A solar cell generates a potential difference of \(0.10 \mathrm{~V}\) when a \(500 \Omega\) resistor is connected across it, and a potential difference of \(0.15 \mathrm{~V}\) when a \(1000 \Omega\) resistor is substituted. What are the (a) internal resistance and (b) emf of the solar cell? (c) The area of the cell is \(5.0 \mathrm{~cm}^{2}\), and the rate per unit area at which it receives energy from light is \(2.0 \mathrm{~mW} / \mathrm{cm}^{2} .\) What is the efficiency of the cell for converting light energy to thermal energy in the 1000 \(\Omega\) external resistor?

Suppose that, while you are sitting in a chair, charge separation between your clothing and the chair puts you at a potential of \(200 \mathrm{~V}\), with the capacitance between you and the chair at \(150 \mathrm{pF}\). When you stand up, the increased separation between your body and the chair decreases the capacitance to \(10 \mathrm{pF}\). (a) What then is the potential of your body? That potential is reduced over time, as the charge on you drains through your body and shoes (you are a capacitor discharging through a resistance). Assume that the resistance along that route is \(300 \mathrm{G} \Omega\). If you touch an electrical component while your potential is greater than \(100 \mathrm{~V}\), you could ruin the component. (b) How long must you wait until your potential reaches the safe level of \(100 \mathrm{~V}\) ? If you wear a conducting wrist strap that is connected to ground, your potential does not increase as much when you stand up; you also discharge more rapidly because the resistance through the grounding connection is much less than through your body and shoes. (c) Suppose that when you stand up, your potential is \(1400 \mathrm{~V}\) and the chair-to-you capacitance is \(10 \mathrm{pF}\). What resistance in that wrist-strap grounding connection will allow you to discharge to \(100 \mathrm{~V}\) in \(0.30 \mathrm{~s}\), which is less time than vou would need to reach for, say, your computer?
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