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Chapter 18: Problem 59

What is the current in a \(60.0-\mathrm{W}\) bulb when connected to a \(120-\mathrm{V}\) emf?

Short Answer

Expert verified
Answer: The current in the circuit is 0.5 A.

Step by step solution

01

1. Identify important formulas

The formula that relates power (P), current (I), and voltage (V) is: P = IV
02

2. Rearrange the formula

To find the current (I), we need to rearrange the formula: I = P/V
03

3. Input given values

We are given P = 60.0 W, and V = 120 V. Plugging these values into the rearranged formula, we get: I = \frac{60.0\,\text{W}}{120\,\text{V}}
04

4. Calculate the current

Now, let's calculate the current (I) by dividing the power by the voltage: I = \frac{60.0}{120} = 0.5\,\text{A}
05

5. Conclusion

The current in the 60.0 W bulb when connected to a 120 V emf is 0.5 A.

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Most popular questions from this chapter

The resistance of a conductor is \(19.8 \Omega\) at \(15.0^{\circ} \mathrm{C}\) and \(25.0 \Omega\) at \(85.0^{\circ} \mathrm{C} .\) What is the temperature coefficient of resistance of the material?
A \(1.5-\mathrm{V}\) flashlight battery can maintain a current of $0.30 \mathrm{A}\( for \)4.0 \mathrm{h}$ before it is exhausted. How much chemical energy is converted to electrical energy in this process? (Assume zero internal resistance of the battery.)
About \(5.0 \times 10^{4} \mathrm{m}\) above Earth's surface, the atmosphere is sufficiently ionized that it behaves as a conductor. The Earth and the ionosphere form a giant spherical capacitor, with the lower atmosphere acting as a leaky dielectric. (a) Find the capacitance \(C\) of the Earth-ionosphere system by treating it as a parallel plate capacitor. Why is it OK to do that? [Hint: Com- pare Earth's radius to the distance between the "plates." (b) The fair-weather electric field is about \(150 \mathrm{V} / \mathrm{m},\) downward. How much energy is stored in this capacitor? (c) Due to radioactivity and cosmic rays, some air molecules are ionized even in fair weather. The resistivity of air is roughly \(3.0 \times 10^{14} \Omega \cdot \mathrm{m}\) Find the resistance of the lower atmosphere and the total current that flows between Earth's surface and the ionosphere. [Hint: since we treat the system as a parallel plate capacitor, treat the atmosphere as a dielectric of uniform thickness between the plates.] (d) If there were no lightning, the capacitor would discharge. In this model, how much time would elapse before Earth's charge were reduced to \(1 \%\) of its normal value? (Thunderstorms are the sources of emf that maintain the charge on this leaky capacitor.)
Capacitors are used in many applications where one needs to supply a short burst of relatively large current. A 100.0 - \(\mu\) F capacitor in an electronic flash lamp supplies a burst of current that dissipates \(20.0 \mathrm{J}\) of energy (as light and heat) in the lamp. (a) To what potential difference must the capacitor initially be charged? (b) What is its initial charge? (c) Approximately what is the resistance of the lamp if the current reaches \(5.0 \%\) of its original value in $2.0 \mathrm{ms} ?$
Find the maximum current that a fully charged D-cell can supply - if only briefly-such that its terminal voltage is at least \(1.0 \mathrm{V}\). Assume an emf of \(1.5 \mathrm{V}\) and an internal resistance of \(0.10 \Omega .\)
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