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A 8 volt battery is connected purely in parallel or in series to three resistors. Two...

A 8 volt battery is connected purely in parallel or in series to three resistors. Two resistors are 9 ohms and 13 Ohms. The current leaving the battery is 17 Amps. What is the power dissipated, in Watts, by the unknown, third resistor?

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Answer #1

I have solved the problem when battery is connected in parallel with the three resistors. For series combination the given value of currentis not valid as it yield potential drop across R1,R2 be 153V and 221V which are not possible for a 8V battery. If you have a valid value of current, to find power dissipated through R3 first you have to solve it for the voltage V3 using the equation( V=V1+V2+V3) I provided. Then use formula for power dissipation i.e., P=IV

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