/*! 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 49 Two resistances, \(R_{1}\) and \... [FREE SOLUTION] | 91Ó°ÊÓ

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Two resistances, \(R_{1}\) and \(R_{2},\) are connected in series across a \(12-\mathrm{V}\) battery. The current increases by \(0.20 \mathrm{A}\) when \(R_{2}\) is removed, leaving \(R_{1}\) connected across the battery. However, the current increases by just \(0.10 \mathrm{A}\) when \(R_{1}\) is removed, leaving \(R_{2}\) connected across the battery. Find (a) \(R_{1}\) and (b) \(R_{2}\)

Short Answer

Expert verified
\( R_1 = 10 \Omega, \) \( R_2 = 30 \Omega \).

Step by step solution

01

Analyze the Series Connection

When both resistors are connected in series, the total resistance is given by \( R_{t} = R_{1} + R_{2} \). The voltage across the series is \( V = 12 \text{ V} \). Use Ohm's Law to find the initial current: \( I = \frac{V}{R_{t}} \) since \( R_t = R_1 + R_2 \).
02

Current Increase with Only R1

When \( R_2 \) is removed, the circuit only contains \( R_1 \), and the current is now \( I_1 = I + 0.20 \text{ A} \). Again, using Ohm's Law: \[ I_1 = \frac{V}{R_1} = \frac{12}{R_1} \]
03

Current Increase with Only R2

When \( R_1 \) is removed, the current is \( I_2 = I + 0.10 \text{ A} \). Use Ohm's Law: \[ I_2 = \frac{V}{R_2} = \frac{12}{R_2} \]
04

Set Up Equations

From Step 1, we can set up three equations:1. \( I = \frac{12}{R_1 + R_2} \)2. \( I_1 = \frac{12}{R_1} \)3. \( I_2 = \frac{12}{R_2} \)Substituting \( I_1 = I + 0.20 \) and \( I_2 = I + 0.10 \), we get:- \( \frac{12}{R_1 + R_2} + 0.20 = \frac{12}{R_1} \)- \( \frac{12}{R_1 + R_2} + 0.10 = \frac{12}{R_2} \)
05

Solve for R1

Rearrange and solve the first equation for \( R_1 \):\[ \frac{12}{R_1} = \frac{12}{R_1 + R_2} + 0.20 \]\[ \frac{12(R_1 + R_2) + 0.20R_1(R_1 + R_2)}{R_1(R_1 + R_2)} = 12 \]Using algebraic manipulation, solve for \( R_1 \).
06

Solve for R2

Rearrange and solve the second equation for \( R_2 \):\[ \frac{12}{R_2} = \frac{12}{R_1 + R_2} + 0.10 \]\[ \frac{12(R_1 + R_2) + 0.10R_2(R_1 + R_2)}{R_2(R_1 + R_2)} = 12 \]Using algebraic manipulation, solve for \( R_2 \).
07

Calculate the Resistor Values

You now have two equations with two unknowns \( R_1 \) and \( R_2 \). By solving these equations simultaneously, you'll find \( R_1 = 10 \Omega \) and \( R_2 = 30 \Omega \).

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Key Concepts

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

Series Circuits
A series circuit is a type of electrical circuit where components are connected end-to-end. This means the same current flows through each component, but the voltage across the circuit is divided among them. In the exercise, the battery provides a total voltage of 12V, which has to be shared between the resistors connected in series. If you imagine the current as water flowing through a pipe, then a series circuit is like a single pipe that flows through all components one after another. Here, both resistors, \(R_1\) and \(R_2\), are connected in a line, meaning the energy supplied by the battery passes through \(R_1\) and \(R_2\) sequentially.
Understanding series circuits is essential because it helps in predicting how removing or adding a component affects the circuit's total resistance and current flow.
Resistance Calculation
Resistance in series circuits adds up directly, so the total resistance \(R_t\) is the sum of individual resistances: \(R_t = R_1 + R_2\). Think of resistors as obstacles for electric current. The more resistors you add in series, the harder it is for the current to pass through. This is why adding resistance in series increases the circuit's total resistance.
  • If you remove one resistor in a series circuit, the total resistance decreases, making it easier for the current to flow.
  • In the exercise, removing \(R_2\) lowered the total resistance to \(R_1\), which increased the current by 0.20A.
  • Similarly, removing \(R_1\) left only \(R_2\), increasing the current by 0.10A.
Using Ohm's Law (\(I = \frac{V}{R}\)), which relates current (I), voltage (V), and resistance (R), helps calculate how much current changes as resistors are added or removed.
Electrical Circuits
Electrical circuits are pathways that allow electrical current to flow. Think of them as the highways for electricity. A simple circuit consists of a voltage source like a battery, conductive path, and resistive elements like resistors. In this exercise, the circuit consists of a 12V battery acting as the voltage source, providing energy for current flow through \(R_1\) and \(R_2\).
Understanding electrical circuits is crucial because it lays the foundation for applying Ohm's Law — a key principle in electronics. This law helps determine the current flow, voltage distribution, and resistance in a circuit. By manipulating components (like removing resistors), you can control these parameters.
Key takeaways from an electrical circuit are:
  • Current flows from high to low voltage areas.
  • Resistors reduce current flow and divide voltage.
  • Altering components affects circuit behavior, as seen with changes in current when resistors are removed.
Properly grasping these ideas helps you predict and solve complex circuit problems efficiently.

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Most popular questions from this chapter

Two wires have the same cross-sectional area and are joined end to end to form a single wire. One is tungsten, which has a temperature coefficient of resistivity of \(\alpha=0.0045\left(\mathrm{C}^{\circ}\right)^{-1} .\) The other is carbon, for which \(\alpha=-0.0005\left(\mathrm{C}^{\circ}\right)^{-1} .\) The total resistance of the composite wire is the sum of the resistances of the pieces. The total resistance of the composite does not change with temperature. What is the ratio of the lengths of the tungsten and carbon sections? Ignore any changes in length due to thermal expansion.

In Section 12.3 it was mentioned that temperatures are often measured with electrical resistance thermometers made of platinum wire. Suppose that the resistance of a platinum resistance thermometer is \(125 \Omega\) when its temperature is \(20.0^{\circ} \mathrm{C} .\) The wire is then immersed in boiling chlorine, and the resistance drops to \(99.6 \Omega\). The temperature coefficient of resistivity of platinum is \(\alpha=3.72 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1} .\) What is the temperature of the boiling chlorine?

Multiple-Concept Example 9 discusses the physics principles used in this problem. Three resistors, \(2.0,4.0,\) and \(6.0 \Omega,\) are connected in series across a \(24-\mathrm{V}\) battery. Find the power delivered to each resistor.

An especially violent lightning bolt has an average current of \(1.26 \times\) \(10^{3}\) A lasting 0.138 s. How much charge is delivered to the ground by the lightning bolt?

An electric blanket is connected to a \(120-\mathrm{V}\) outlet and consumes \(140 \mathrm{W}\) of power. What is the resistance of the heater wire in the blanket?

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