V battery as a 1.5 V emf in series with a resistor known as the \"internal resis

Question

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 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 3.0 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

current in circuit = 1.5 /(4) = 0.375 amps total power dissipated in circuit = I^2 R = 0.5625 watt power dissipated in internal resistance = 0.141 watt required ratio =0.141 watt /0.5625 watt =0.25 or 25 %

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