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Chapter 18: Problem 69

A battery has a \(6.00-\mathrm{V}\) emf and an internal resistance of $0.600 \Omega$. (a) What is the voltage across its terminals when the current drawn from the battery is \(1.20 \mathrm{A} ?\) (b) What is the power supplied by the battery?

Short Answer

Expert verified
Answer: The terminal voltage is 5.28 V, and the power supplied by the battery is 6.336 W.

Step by step solution

01

Identify the given parameters

The given parameters are: - emf (E) = 6.00 V - Internal resistance (r) = 0.600 Ω - Current (I) = 1.20 A
02

Find the terminal voltage V

We can find the terminal voltage (V) by using the relationship between emf, internal resistance, and the current drawn from the battery: V = E - Ir where E = emf I = current r = internal resistance Now, we can plug in the given values: V = 6.00 V - (1.20 A)(0.600 Ω) = 6.00 V - 0.720 V = 5.28 V Thus, the terminal voltage when the current is drawn is 5.28 V.
03

Calculate the power supplied by the battery

Now, we can calculate the power supplied by the battery using the formula: Power (P) = VI where V = terminal voltage I = current Plugging in the values: P = (5.28 V)(1.20 A) = 6.336 W Therefore, the power supplied by the battery is 6.336 W.

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Most popular questions from this chapter

A wire with cross-sectional area \(A\) carries a current \(I\) Show that the electric field strength \(E\) in the wire is proportional to the current per unit area \((I / A)\) and identify the constant of proportionality. [Hint: Assume a length \(L\) of wire. How is the potential difference across the wire related to the electric field in the wire? (Which is uniform?) Use \(V=I R\) and the connection between resistance and resistivity.]
A copper wire and an aluminum wire of the same length have the same resistance. What is the ratio of the diameter of the copper wire to that of the aluminum wire?
A silver wire of diameter 1.0 mm carries a current of \(150 \mathrm{mA} .\) The density of conduction electrons in silver is $5.8 \times 10^{28} \mathrm{m}^{-3} .$ How long (on average) does it take for a conduction electron to move \(1.0 \mathrm{cm}\) along the wire?
An aluminum wire of diameter 2.6 mm carries a current of 12 A. How long on average does it take an electron to move \(12 \mathrm{m}\) along the wire? Assume 3.5 conduction electrons per aluminum atom. The mass density of aluminum is \(2.7 \mathrm{g} / \mathrm{cm}^{3}\) and its atomic mass is $27 \mathrm{g} / \mathrm{mol}$.
Near Earth's surface the air contains both negative and positive ions, due to radioactivity in the soil and cosmic rays from space. As a simplified model, assume there are 600.0 singly charged positive ions per \(\mathrm{cm}^{3}\) and 500.0 singly charged negative ions per \(\mathrm{cm}^{3} ;\) ignore the presence of multiply charged ions. The electric field is $100.0 \mathrm{V} / \mathrm{m},$ directed downward. (a) In which direction do the positive ions move? The negative ions? (b) What is the direction of the current due to these ions? (c) The measured resistivity of the air in the region is $4.0 \times 10^{13} \Omega \cdot \mathrm{m} .$ Calculate the drift speed of the ions, assuming it to be the same for positive and negative ions. [Hint: Consider a vertical tube of air of length \(L\) and cross-sectional area \(A\) How is the potential difference across the tube related to the electric field strength?] (d) If these conditions existed over the entire surface of the Earth, what is the total current due to the movement of ions in the air?
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