In a simple electric circuit, ohm's law states that v=irv=ir, where vv is the voltage in volts, ii is the current in amperes, and rr is the resistance in ohms. assume that, as the battery wears out, the voltage decreases at 0.030.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.040.04 ohms per second. when the resistance is 200200 ohms and the current is 0.040.04 amperes, at what rate is the current changing?

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3 September, 21:08

In a simple electric circuit, ohm's law states that v=irv=ir, where vv is the voltage in volts, ii is the current in amperes, and rr is the resistance in ohms. assume that, as the battery wears out, the voltage decreases at 0.030.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.040.04 ohms per second. when the resistance is 200200 ohms and the current is 0.040.04 amperes, at what rate is the current changing?

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Annabelle · Answer 1

We take the derivative of Ohm's law with respect to time: V = IR

Using the product rule:

dV/dt = I (dR/dt) + R (dI/dt)

We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:

-0.03 V/s = (0.04 A) (0.04 ohm/s) + (200 ohms) (dI/dt)

dI/dt = - 0.000158 = - 1.58 x 10^-4 A/s