Ch. 26 - DC Circuits

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo

Giancoli Douglas 5th edition

Ch. 26 - DC Circuits

Problema 35

Capitolo 25, Problema 35

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.

Guida verificata passo dopo passo

Start by recalling the formula for power dissipated in a resistor: \( P = \frac{V^2}{R} \), where \( V \) is the voltage across the resistor and \( R \) is its resistance.

For the parallel configuration, the equivalent resistance \( R_{\text{eq,parallel}} \) of \( n \) identical resistors (each with resistance \( R \)) is given by \( \frac{1}{R_{\text{eq,parallel}}} = \frac{1}{R} + \frac{1}{R} + \dots + \frac{1}{R} = \frac{n}{R} \), so \( R_{\text{eq,parallel}} = \frac{R}{n} \). The power dissipated in this case is \( P_{\text{parallel}} = \frac{V^2}{R_{\text{eq,parallel}}} = \frac{V^2}{R/n} = n \frac{V^2}{R} \).

For the series configuration, the equivalent resistance \( R_{\text{eq,series}} \) of \( n \) identical resistors is \( R_{\text{eq,series}} = R + R + \dots + R = nR \). The power dissipated in this case is \( P_{\text{series}} = \frac{V^2}{R_{\text{eq,series}}} = \frac{V^2}{nR} \).

To find the factor by which the power is decreased, calculate the ratio \( \frac{P_{\text{series}}}{P_{\text{parallel}}} \). Substituting the expressions for \( P_{\text{series}} \) and \( P_{\text{parallel}} \), we get \( \frac{P_{\text{series}}}{P_{\text{parallel}}} = \frac{\frac{V^2}{nR}}{n \frac{V^2}{R}} = \frac{1}{n^2} \).

Thus, the power transformed in the series configuration is decreased by a factor of \( n^2 \) compared to the parallel configuration.

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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 as V = IR, which is fundamental in analyzing electrical circuits, including those with resistors in series and parallel.

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03:07

Resistance and Ohm's Law

Power in Electrical Circuits

The power (P) consumed in an electrical circuit is defined as the rate at which energy is used or converted, calculated using the formula P = IV, where I is the current and V is the voltage. In resistive circuits, power can also be expressed as P = I²R or P = V²/R, depending on the configuration of the resistors, which is crucial for understanding how power changes with different connections.

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Power in Circuits

Resistor Configurations: Series vs. Parallel

In a series configuration, resistors are connected end-to-end, resulting in a total resistance that is the sum of individual resistances (R_total = R1 + R2 + ... + Rn). In contrast, in a parallel configuration, the total resistance is reduced and calculated using the formula 1/R_total = 1/R1 + 1/R2 + ... + 1/Rn. This difference in resistance affects the current and power distribution in the circuit, leading to the conclusion that power is decreased by a factor of n² when switching from parallel to series.

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Combining Resistors in Series & Parallel

Pratica correlata

Domanda del libro di testo

(III) Determine the net resistance in Fig. 26–61 (a) between points a and c, and (b) between points a and b. Assume R' ≠ R. [Hint: Apply an emf between the two points in each case and determine currents; use symmetry at junctions.]

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Domanda del libro di testo

(II) Suppose two batteries, with unequal emfs of 2.00 V and 3.00 V, are connected as shown in Fig. 26–63. If each internal resistance is r = 0.350Ω and R = 4.00Ω, what is the voltage across the resistor R?

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Domanda del libro di testo

(III) (a) Determine the currents I₁, I₂, and I₃ in Fig. 26–58. Assume the internal resistance of each battery is r = 1.0 Ω.

(b) What is the terminal voltage of the 6.0-V battery?

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Domanda del libro di testo

(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?

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Domanda del libro di testo

(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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Domanda del libro di testo

(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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