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Ch. 26 - DC Circuits
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 26 - DC CircuitsProblem 26.39
Chapter 25, Problem 26.39

(III) If the 25-Ω resistor in Fig. 26–59 is shorted out (resistance = 0 ), what then would be the current through the 15-Ω resistor?

Verified step by step guidance
1
Identify the resistors in the circuit and their arrangement. Since the 25-Ω resistor is shorted out, it effectively has no resistance and does not affect the circuit.
Determine the total resistance in the circuit. With the 25-Ω resistor shorted, focus on the remaining resistors. If they are in series, add their resistances directly. If in parallel, use the formula for parallel resistors: \(\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + ...\).
Calculate the total voltage supplied in the circuit if not given, assume a standard value or use the given battery voltage.
Apply Ohm's Law, \(V = IR\), to find the total current in the circuit using the total resistance calculated and the total voltage.
To find the current through the 15-Ω resistor specifically, use the current division rule if the resistors are in parallel, or note that the current is the same through each resistor if they are in series.

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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 I = V/R. Understanding this law is crucial for analyzing circuits and determining how current behaves when resistances change.
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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 in a parallel circuit, the voltage across each component is the same. Knowing how to identify these configurations helps in calculating the total resistance and current flow when components are added or removed, such as shorting a resistor.
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Equivalent Resistance

Equivalent resistance is the total resistance of a circuit when multiple resistors are combined. For resistors in series, the equivalent resistance is the sum of all resistances, while for resistors in parallel, the equivalent resistance can be found using the formula 1/R_eq = 1/R1 + 1/R2 + ... + 1/Rn. This concept is essential for simplifying complex circuits and determining how changes, like shorting a resistor, affect overall current flow.
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Related Practice
Textbook Question

[In these Problems neglect the internal resistance of a battery unless the Problem refers to it.]


(II) A power supply has a fixed output voltage of 12.0 V, but you need VT = 3.0 V output for an experiment. (a) Using the voltage divider shown in Fig. 26–47, what should R₂ be if R₁ is 16.5 Ω? (b) What will the terminal voltage VT be if you connect a load to the 3.0-V output, assuming the load has a resistance of 7.0Ω?

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Textbook Question

A voltage V is applied to n identical resistors connected in parallel. If the resistors are instead all connected in series with the applied voltage, show that the power transformed is decreased by a factor n².

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Textbook Question

(II) Determine the magnitudes and directions of the currents in each resistor shown in Fig. 26–57. The batteries have emfs of ε1 = 9.0V and ε2 = 12.0V and the resistors have values of R1 = 25 Ω, R2 = 48 Ω, and R3 = 35 Ω.

(a) Ignore internal resistance of the batteries.

(b) Assume each battery has internal resistance r = 1.0 Ω.

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Textbook Question

[In these Problems neglect the internal resistance of a battery unless the Problem refers to it.]


(III) You are designing a wire resistance heater to heat an enclosed container of gas. For the apparatus to function properly, this heater must transfer heat to the gas at a very constant rate. While in operation, the resistance of the heater will always be close to the value R = R₀, but may fluctuate slightly causing its resistance to vary a small amount ∆R ( << R₀ ). To maintain the heater at constant power, you design the circuit shown in Fig. 26–50, which includes two resistors, each of resistance R′. Determine the value for R′ so that the heater power P will remain constant even if its resistance R fluctuates by a small amount. [Hint: If ∆R << R₀ , then ΔP≈ΔRdPdR∣R=R0\(\Delta\) P\(\approx\) \(\Delta\) R\(\left\). \(\frac{dP}{dR}\]\right\)|_{R=R_{0}}]

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Textbook Question

(II) (a) What is the potential difference between points a and d in Fig. 26–55 (similar to Fig. 26–12, Example 26–8), and (b) what is the terminal voltage of each battery?

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Textbook Question

[In these Problems neglect the internal resistance of a battery unless the Problem refers to it.]


(II) What is the net resistance of the circuit connected to the battery in Fig. 26–46?

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