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Ch 26: Direct-Current Circuits
All textbooksYoung & Freedman Calc 14th EditionCh 26: Direct-Current CircuitsProblem 26
Chapter 26, Problem 26

The batteries shown in the circuit in Fig. E26.24 have negligibly small internal resistances. Circuit diagram showing two batteries and resistors R1 and R2 for Kirchhoff's Loop Rule.
Find the current through (a) the 30.0-Ω resistor.

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1
Identify the components in the circuit: two batteries (E1 and E2) and two resistors (R1 and R2).
Apply Kirchhoff's Voltage Law (KVL) to the loop. This law states that the sum of all voltages around a closed loop must equal zero.
Write the KVL equation for the loop: E1 - I*R1 - E2 - I*R2 = 0, where I is the current through the circuit.
Rearrange the equation to solve for the current I: I = (E1 - E2) / (R1 + R2).
Substitute the given values for E1, E2, R1, and R2 into the equation to find the current I through the 30.0-Ω resistor (R2).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ohm's Law

Ohm's Law states that the current (I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points and inversely proportional to the resistance (R) of the conductor. This relationship is expressed mathematically as V = IR. Understanding this law is crucial for analyzing circuits, as it allows us to calculate the current through resistors when the voltage and resistance are known.
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Resistance and Ohm's Law

Kirchhoff's Loop Rule

Kirchhoff's Loop Rule, also known as the voltage law, states that the sum of the electrical potential differences (voltage) around any closed circuit loop must equal zero. This principle is essential for analyzing complex circuits, as it helps determine the voltage drops across components and ensures that energy is conserved within the circuit. Applying this rule allows us to set up equations to solve for unknown currents and voltages.
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Intro to Kirchhoff's Loop Rule

Series and Parallel Circuits

In electrical circuits, components can be arranged in series or parallel configurations. In a series circuit, the same current flows through all components, and the total resistance is the sum of individual resistances. In contrast, in a parallel circuit, the voltage across each component is the same, and the total current is the sum of the currents through each branch. Understanding these configurations is vital for analyzing how current flows through the resistors in the given circuit.
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Related Practice
Textbook Question
Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. The two light bulbs are connected in series across a 120-V line. Afterwards, the two light bulbs are connected in parallel across the 120-V line. (g) In each situation, which of the two bulbs glows the brightest?
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Textbook Question
Light Bulbs in Series and in Parallel. Two light bulbs have constant resistances of 400Ω and 800Ω. The two light bulbs are connected in series across a 120-V line. Afterwards, the two light bulbs are connected in parallel across the 120-V line. (h) In which situation is there a greater total light output from both bulbs combined?
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Textbook Question
In the circuit shown in Fig. E26.31 the batteries have negligible internal resistance and the meters are both idealized. With the switch S open, the voltmeter reads 15.0 V.

(b) What will the ammeter read when the switch is closed?
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Textbook Question
In the circuit shown in Fig. E26.27

find (b) the unknown emfs ε_1 and ε_2.
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Textbook Question
In the circuit shown in Fig. E26.33 all meters are idealized and the batteries have no appreciable internal resistance.

(a) Find the reading of the voltmeter with the switch S open. Which point is at a higher potential: a or b?
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Textbook Question
An emf source with ε = 120V, a resistor with R = 80.0Ω, and a capacitor with C = 4.00 μF are connected in series. As the capacitor charges, when the current in the resistor is 0.900 A, what is the magnitude of the charge on each plate of the capacitor?

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