Answer to Question #14400 in Algebra for john
By observing the rate of increase of the temperature in the water bath, it is determined that the power dissipated in the resistor is 11.0W.
Assuming that the voltage across the resistor is constant, what is the voltage v (in Volts) across the resistor?
What is the current i (in Amperes) entering the network N from the resistor?
P = V * I = V^2 / R = R * I^2
V = sqrt(P * R) =
sqrt(11.0 * 8.0) = 9.381 V.
I = V / R = 9.381 / 8.0 = 1.173 A.
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