Ch. 25 - Electric Current and Resistance

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 25 - Electric Current and Resistance

Problem 46

Chapter 24, Problem 46

Calculate the peak current in a 2.5-k Ω resistor connected to a 220-V rms ac source.

Verified step by step guidance

Step 1: Understand the given values and the formula to use. The problem provides the resistance $ R = 2.5 \; \text{k}\Omega = 2500 \; \Omega $ and the root mean square (rms) voltage $ V_{\text{rms}} = 220 \; \text{V} $. The goal is to calculate the peak current $ I_{\text{peak}} $.

Step 2: Recall the relationship between rms voltage and peak voltage. The peak voltage $ V_{\text{peak}} $ is related to the rms voltage by the equation: $ V_{\text{peak}} = \sqrt{2} \cdot V_{\text{rms}} $.

Step 3: Use Ohm's Law to relate the peak current to the peak voltage and resistance. Ohm's Law states $ I = \frac{V}{R} $. For peak current, this becomes $ I_{\text{peak}} = \frac{V_{\text{peak}}}{R} $.

Step 4: Substitute $ V_{\text{peak}} = \sqrt{2} \cdot V_{\text{rms}} $ into the equation for $ I_{\text{peak}} $. This gives $ I_{\text{peak}} = \frac{\sqrt{2} \cdot V_{\text{rms}}}{R} $.

Step 5: Plug in the given values $ V_{\text{rms}} = 220 \; \text{V} $ and $ R = 2500 \; \Omega $ into the equation $ I_{\text{peak}} = \frac{\sqrt{2} \cdot 220}{2500} $. Simplify the expression to find the peak current.

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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. It is mathematically expressed as I = V/R. This fundamental principle is essential for calculating current in electrical circuits.

RMS Voltage

RMS (Root Mean Square) voltage is a way of expressing the effective value of an alternating current (AC) voltage. It represents the equivalent direct current (DC) voltage that would deliver the same power to a resistor. For a sinusoidal AC source, the RMS voltage is approximately 0.707 times the peak voltage, making it crucial for calculations involving AC circuits.

Peak Current

Peak current refers to the maximum instantaneous current that flows through a circuit during one cycle of an AC waveform. It can be calculated from the RMS current using the relationship I_peak = I_rms × √2. Understanding peak current is important for analyzing the performance and safety of electrical components in AC circuits.

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A flashlight uses two AA 1.5-V batteries connected in series to provide 3.0 V across the bulb, as in Fig. 25–4b. The bulb draws 135 mA when turned on.

(a) Calculate the resistance of the bulb and the power dissipated.

(b) By what factor would the power increase if four AA batteries in series (total 6.0 V) were used with the same bulb? (Neglect heating effects of the filament.) Why shouldn’t you try this?

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