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Home Q & A Board Science-Related Homework Help Physics
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rkarthik301 rkarthik301
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11 years ago
For the real battery shown in the figure, calculate the power dissipated by a resistor R connected to the battery when R1 = 2.3 \Omega.

The figure shows a better with 10 omega on top and 1.5 volts on bottom.
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#1
11 years ago
The internal resistance of the battery is 10 Ohm. So the total resistance of the (serial) circuit is 12.3 Ohm.
Therefore the current in the circuit will be:

I = V / R = 1.5 V / 12.3 Ohm = 0.122 Ampere.

Hence the power dissipated in the resistance is:

P = I^2 R = (0.122 A)^2 * 2.3 Ohm = 34.2 mW
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