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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 惟 - 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 惟) = 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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