http://session.masteringphysics.com/problemAsset/1072293/2/25.1e.jpg The emf and

Question

http://session.masteringphysics.com/problemAsset/1072293/2/25.1e.jpg

The emf and the internal resistance of a battery are as shown. In Fig. 25.1e, the power dissipated within the battery is 9.5 W when a resistor is connected across the terminals. The power dissipated by the resistor is closest to:

120 W
240 W
180 W
300 W
350 W

I have been struggling for hours on how to solve this. Am I supposed to find the current? If so, wouldn't I need to know the internal resistance of the battery to find it? Is this even a circuit?

Explanation / Answer

   Let the EMF of battery be E, internal resistance be r, external resistance connected across battery terminals be R and current be I, then    E   =   I * R   +   I * r         ----------------- (1)    Given   E   =   97.0   V                r   =   5.0   O    Power dissipiated in battery   PB   =   I2 * r    9.5   =   I2 * 5    Current   I   =   v(9.5/5)                      =   1.378   A    Using equation (1)    97.0   =   1.378 * R   +   1.378 * 5    R   =   (97.0   -   6.89) / 1.378          =   65.39   O    Power dissipiated in external resistor   P   =   I2 * R                                                                   =   1.3782 * 65.39                                                                   =   124.17   W    That is closest to 120   W.        

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