V=1.1V Just need help with part B A real battery is not just an emf. We can mode
ID: 1423906 • Letter: V
Question
V=1.1V
Just need help with part B
A real battery is not just an emf. We can model a real 1.5 V battery as a 1.5 V emf in series with a resistor known as the "internal resistance", as shown in the figure. A typical battery has 1.0 Ohm internal resistance due to imperfections that limit current through the battery. When there's no current through the battery, and thus no voltage drop across the internal resistance, the potential difference between its terminals is 1.5 V, the value of the emf. Suppose the terminals of this battery are connected to a 2.5 Ohm resistor. What is the potential difference between the terminals of the battery? Express your answer using two significant figures. What fraction of the battery's power is dissipated by the internal resistance? Express your answer using two significant figures.Explanation / Answer
As V=1.1V
Power dissipated = I ² x R = .428 ² x 1 = 0.183 watts
Power supplied = I ² x R(total)
= .183 ² x 3.5 = 0.117 W
So 0.183 / 0.117 is the fraction of power dissipated in the internal resistance
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