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

Problem 23.55 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 2 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 2.9 2 resistor. Figure 1 of 1 1.0 1.5 V Part A What is the potential difference between the terminals of the battery? Express your answer using two significant figures V 1.0 Submit My Answers Give Up incorrect; Try Again; 4 attempts remaining Part B What fraction of the battery's power is dissipated by the internal resistance? Express your answer using two significant figures AP Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

As per the described arrangement, the internal resistance of 1 ohm is in series with the 2.9 ohm resistor.

Req = 1 + 2.9 - 3.9 Ohm

from Ohm/s law, V = IR =>I = V/R

I = 1.5/3.9 = 0.385 A

Drop across the r will be:

Vt = r I = 1 x 0.385 = 0.385 Volts

Vt = 1.5 - .385 = 1.115

Hence, Vt = 1.115 Volts

b)Pt = V I = 0.385 x 0.385 = 0.148 W

Pr = I^2 R = 0.385^2 x 3.9 = 0.578 W

deltaP/P = 0.148/0.578 = 0.256

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