Three resistors are connected in series across a battery. The value of each resi

Question

Three resistors are connected in series across a battery. The value of each resistance and its maximum power rating are as follows: 6.6? and 22.4 W, 42.6? and 12.6 W, and 19.6? and 11.1 W. (a) What is the greatest voltage that the battery can have without one of the resistors burning up? (b) How much power does the battery deliver to the circuit in (a)?

Explanation / Answer

Power dissipated by a resistor is calculated using P = I*I*R So the first step is to determine what the maximum allowed current would be for each resistor. 22.4 W = I^2*6.6 => I = 1.842 amps 12.6 = I^2*42.6 => I = 0.544 amps 11.1 = I^2*19.6 => I = 0.753 amps. Now since all three resistors are in series, the current flowing through one resistor will be the same for all three resistors. The lowest maximum current is for the 42.6 ohm resistor at 0.544 amps. So the voltage must not force more than 0.544 amps otherwise the 42.6 ohm resistor will burn up. The total resistance of the circuit is the sum of all three resistors assuming an ideal battery. So R-total = 6.6 + 42.6 + 19.6 = 68.8 ohms. E = I*R so the maximum battery voltage is E = .544*68.8 = 37.4272 Volts Power delivered by battery = I^2*R = 0.544^2 *68.8 = 20.36 W

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