The 10.00 V battery in Fig. E26.28 is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. Find the current in each branch.
Ch 26: Direct-Current Circuits
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 26, Problem 27a
In the circuit shown in Fig. E26.27 find the current in the 3.00 Ω resistor.

Verified step by step guidance1
Step 1: Analyze the circuit diagram. The circuit contains two batteries (E₁ and E₂), resistors (4.00 Ω, 3.00 Ω, 6.00 Ω, and R), and currents labeled in different branches. The goal is to find the current through the 3.00 Ω resistor.
Step 2: Apply Kirchhoff's Current Law (KCL) at the junctions. KCL states that the sum of currents entering a junction equals the sum of currents leaving the junction. Use the given currents (2.00 A, 3.00 A, and 5.00 A) to determine the current through the 3.00 Ω resistor.
Step 3: Apply Kirchhoff's Voltage Law (KVL) to the loops in the circuit. KVL states that the sum of the potential differences (voltage drops and rises) around any closed loop is zero. Write equations for the loops involving E₁, E₂, R, and the resistors.
Step 4: Use Ohm's Law (V = IR) to relate the voltage across the 3.00 Ω resistor to the current through it. Substitute the current values and resistances into the equations derived from KVL to solve for the unknowns.
Step 5: Solve the system of equations obtained from KCL and KVL to find the current through the 3.00 Ω resistor. Ensure consistency with the given currents and resistances in the circuit.

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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 voltage and resistance values are known.
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Series and Parallel Circuits
In electrical circuits, components can be arranged in series or parallel configurations. In a series circuit, the current is the same through all components, while the total voltage is the sum of the individual voltages. In a parallel circuit, the voltage across each component is the same, but the total current is the sum of the currents through each branch. Recognizing the configuration of resistors in the circuit is essential for calculating the total resistance and current.
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Combining Capacitors in Series & Parallel
Kirchhoff's Laws
Kirchhoff's Laws consist of two fundamental principles for analyzing electrical circuits: Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL). KCL states that the total current entering a junction equals the total current leaving it, while KVL states that the sum of the electrical potential differences (voltage) around any closed circuit loop must equal zero. These laws are vital for solving complex circuits and determining unknown currents and voltages.
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Kirchhoff's Junction Rule
Related Practice
Textbook Question
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Textbook Question
The 5.00 V battery in Fig. E26.28 is removed from the circuit and replaced by a 15.00 V battery, with its negative terminal next to point b. The rest of the circuit is as shown in the figure. Find the current in each branch.
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Textbook Question
The batteries shown in the circuit in Fig. E26.24 have negligibly small internal resistances. Find the current through the 30.0-Ω resistor.
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Textbook Question
In the circuit shown in Fig. E26.23, ammeter A1 reads 10.0 A and the batteries have no appreciable internal resistance. What is the resistance of R?
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Textbook Question
Find the emfs and in the circuit of Fig. E26.26, and find the potential difference of point relative to point .
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Textbook Question
In the circuit shown in Fig. E26.27 find the unknown emfs and .
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