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# Answer to Question #14400 in Algebra for john

Question #14400
The picture shows a resistor connected to some unknown network N. The resistor is immersed in an isolated water bath, and its temperature is observed and recorded. The resistor has resistance R=8.0&Omega;. 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? unanswered What is the current i (in Amperes) entering the network N from the resistor? unanswered
1
2012-09-07T11:13:58-0400
R = 8.0 &Omega;, P = 11.0 W.
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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