V=IR If the current in an electrical system is dI/dt=-8A/s while voltage=10V & I=56A, rate of R=??

Question

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Electrical Systems are governed by Ohm's law, which states the V=IR, where V=voltage, I=current, and R=resistance. If the current in an electrical system is decreasing at a rate of 8 A/s while the voltage remains constant at 10V, at waht rate is the resistance increasing when the current is 56 A? Please provide steps/explanation. THANK YOU!

Answer 1

The resistance is increasing at .1785 ohms!!!

Just use the ohm's law

10/56=.1785 that's the resistance

then multiply the resistance and the current=volts

.1785 x 56 =9.99999volts

This is the result.

Hope this help!!!

Answer 2

Take the equation V = IR and solve for R so it looks like R = V/I then take the derivative dI/dt of this equation should get

dR/dt = V * ln( I) * dI/dt if I did this right then plug in V and I and dI/dt which you know to get dR/dt .

I hope this is right.