Yn, IR = (2.0 A) (4.0 ) = 8.0 V Considering them as the terminals of the source,

Question

Yn, IR = (2.0 A) (4.0 ) = 8.0 V Considering them as the terminals of the source, we have Using a source with an emf of 12 V and an internal resistance r of 2.0 , we add a 4.0 resistor to form the complete circuit shown in (Figure 1). We will allow a current to flow in order to see how the voltmeter and ammeter readings change. What are the ammeter and voltmeter readings now? VabEIr 12 V- (2.0 A) (2.0 2) 8.0 V Either way, we conclude that the voltmeter reads Vab 8.0 V REFLECT The terminal voltage Vab of the battery is less than the battery emf because of the potential drop across the battery's internal resistance T Part A - Practice Problem: If a different resistor R is used and the voltmeter reads 6.1 V, what is the ammeter reading? Express your answer in amperes to two significant figures Figure 1 of 1 Submit uest Answer A I Provide Feedback Next >

Explanation / Answer

Voltmeter reading is equal to the terminal voltage of battery,

V_terminal = e - I r

6.1 = 12 - 2I

I = 2.95 A  

and I = e / (R + r)

R + 2 = 12/2.95

R = 2.07 A ......Ans OR 2.1 A

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