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Homework Statement
You are given a battery of 220 V and a radio device that requires 1600 W to run. You are also given four resistors: three 3 Ω resistors and one 20 Ω resistor. What circuit configuration will allow the radio device to run without getting fried?
The Attempt at a Solution
Power = (Voltage)2/Resistance
1600 W = (220 V)2/R
R = 30.25 Ω
So this device would require 30.25 Ω to function correctly. However, based on several circuit configurations, this magnitude of resistance isn't obtainable? For example, if I used just a complete series circuit, then the total resistance would be:
R total = R1 + R2 + R3 + R4
R total = 3 + 3 + 3 + 20 = 29 Ω total which isn't exactly 30.25
Incorporating a parallel circuit would cause resistance values to be even less than their original values because parallel circuits require the reciprocal resistance values. So if I had a circuit consisting of all the resistors parallel, then I would get:
1/R total = 1/R1 + 1/R2 + 1/R3 + 1/R4
1/R total = 1/3 + 1/3 + 1/3 + 1/20
1/R total = 0.95 Ω, which is way less
Thus, can someone point me in the right direction?