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A battery has an emf of 12 V and connected to a resistor of 3Omega. The current in the circuit is 3.93 A. Calculate (a) terminal voltage and the internal resistance of the battery (b) power delivered by the battery and power delivered to the resistor. |
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Answer» Solution :The given VALUES `I=3.93A, xi = 12 V, R=3Omega` (a) The terminal voltage of the battery is equal to voltage drop ACROSS the resistor `V=IR=3.93xx3=11.79V` The internal resistance of the battery, `r=[(xi-V)/(V)]R=[(12-11.79)/(11.79)]xx3=0.05Omega` (b) The power delivered by the battery `P=I xi=3.93xx12=47.1W` The power delivered to the resistor `=I^(2)R=46.3W` The REMAINING power `=(47.1-46.3)P=0.772W` is delivered to the internal resistance and cannot be used to do useful work. (it is equal to `I^(2)r`). |
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