A real battery is not just an emf. We can model a real 1.5 V battery as a 1.5 V

Question

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(Figure 1). A typical battery has 1.0 omega internal resistance due to imperfections that limit current through the battery. When there's is no current through the battery, and thus no voltage drop across the internal resistance, the potential difference between its terminal is 1.5 V, the value of the emf. Suppose the terminal of this battery are connected to a 2.5 Omega resistor. What are the potential difference between the terminal of the battery? Express your answer using two significant figures. V=1.1 V What fractions of the battery's power is dissipated by the internal resistance? Express your answer using two significant figures.

Explanation / Answer

I see you've correctly calculated the emf of the battery, quickly we write the equations

V= I R

R = 2,5 +1 = 3,5

I = V/R

I = 1,5/ 3,5 = 0.43 A

Let's concentrate on the power dissipated

P= I V

We calculate the power of internal resistance

P = 0.43 0,4 Pint = 0.172 W

The total power

Pt = 0,43 1,5 Pt= 0.645 W

Pint/Pt = 0.172/0.645

Pint/Pt = 0.26

Pint/Pt % = 0.26 100

Pint/Pt % = 26%

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