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

Question

A real battery is not just an emf. We can model a real 1.5v battery as a 1.5v emf in series with a resistor known as the "internal resistance", as shown in the figure(Figure 1) . A typical battery has 1.0 ohms 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 1.5v , the value of the emf. Suppose the terminals of this battery are connected to a 2.7 ohm resistor. What is the potential difference between the terminals of the battery? What fraction of the battery's power is dissipated by the internal resistance?

Explanation / Answer

Mr. Uncouth used a voltage-divider principle to which you probably haven't been exposed. You can find your answers directly using Ohms Law alone: Total resistance in series: 1 + 2.7 = 3.7 ohm Current flowing through all resistors: I=V/R= 1.5V / 3.7ohm = 0.4054A The voltage drop across the terminals of the battery is equal to the voltage drop across the 3 ohm resistor. That drop is: V=IxR = 0.4054A x 2.7ohm = 1.09459V Answer: 1.09459V

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