An idealized voltmeter is connected across the terminals of a 18.0- V battery, a

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An idealized voltmeter is connected across the terminals of a 18.0-V battery, and a 80.0-?appliance is also connected across its terminals. If the voltmeter reads 10.0V . Part A How much power is being dissipated by the appliance? P =   W   SubmitMy AnswersGive Up Part B What is the internal resistance of the battery? r =   ?   SubmitMy AnswersGive Up An idealized voltmeter is connected across the terminals of a 18.0-V battery, and a 80.0-?appliance is also connected across its terminals. If the voltmeter reads 10.0V . An idealized voltmeter is connected across the terminals of a 18.0-V battery, and a 80.0-?appliance is also connected across its terminals. If the voltmeter reads 10.0V . An idealized voltmeter is connected across the terminals of a 18.0-V battery, and a 80.0-?appliance is also connected across its terminals. If the voltmeter reads 10.0V . An idealized voltmeter is connected across the terminals of a 18.0-V battery, and a 80.0-?appliance is also connected across its terminals. If the voltmeter reads 10.0V . Part A How much power is being dissipated by the appliance? P =   W   SubmitMy AnswersGive Up Part B What is the internal resistance of the battery? r =   ?   SubmitMy AnswersGive Up Part A How much power is being dissipated by the appliance? P =   W   SubmitMy AnswersGive Up Part B What is the internal resistance of the battery? r =   ?   SubmitMy AnswersGive Up Part A How much power is being dissipated by the appliance? P =   W   SubmitMy AnswersGive Up Part B What is the internal resistance of the battery? r =   ?   SubmitMy AnswersGive Up Part A How much power is being dissipated by the appliance? P =   W   SubmitMy AnswersGive Up Part B What is the internal resistance of the battery? r =   ?   SubmitMy AnswersGive Up Part A How much power is being dissipated by the appliance? P =   W   SubmitMy AnswersGive Up Part A How much power is being dissipated by the appliance? P =   W   SubmitMy AnswersGive Up P =   W   P =   W   SubmitMy AnswersGive Up Part B What is the internal resistance of the battery? r =   ?   SubmitMy AnswersGive Up Part B What is the internal resistance of the battery? r =   ?   SubmitMy AnswersGive Up r =   ?   r =   ?   SubmitMy AnswersGive Up An idealized voltmeter is connected across the terminals of a 18.0-V battery, and a 80.0-?appliance is also connected across its terminals. If the voltmeter reads 10.0V . Part A How much power is being dissipated by the appliance? P =   W   SubmitMy AnswersGive Up Part B What is the internal resistance of the battery? r =   ?   SubmitMy AnswersGive Up

Explanation / Answer

A) P = V^2/R
So P = (10^2)/80
P = 1.25 W

B) Well, the voltage taken away due to the internal resistance can be added to the voltage across the appliance and that would be 18 V. Also, V = IR
Therefore;
18.0 V = I(80 ohms) + Ir (r = internal resistance)
So we need to find I using V=IR for the appliance specifically.
I = V/R
I = 10/80 =0.125 A
So now we have:
18.0 V = (0.125 A) (80.0 ohms) + (0.125 A)r (the internal resistance is in series with any resistors in the circuit, so they have the same current)
18.0 V = 10 V + (0.125A)r
8 V = (0.125 A)r
r = 64 ohms

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