Question

The voltage v in a simple electrical circuit is slowly decreas- ing as the battery wears out. the resistance r is slowly increasing as the resistor heats up. use ohm’s law, v − ir, to find how the current i is changing at the moment when r − 400 v, i − 0.08a, dvydt − 20.01vys, and drydt − 0.03 vys.

Answer 1

According to Ohm's law, the voltage drop V across a resistor when a current flows through it is calculated using the equation V = IR, where I equals the current in amps (A) and R is the resistance in ohms (Ω). The rate of change of voltage with respect to time is given by dV/dt. Similarly, the rate of change of resistance with respect to time is given by dR/dt. To find dI/dt, we differentiate the Ohm's law equation V = IR with respect to time and solve for dI/dt.

Step-by-step explanation:

According to Ohm's law, the voltage drop V across a resistor when a current flows through it is calculated using the equation V = IR, where I equals the current in amps (A) and R is the resistance in ohms (Ω). The rate of change of voltage with respect to time is given by dV/dt. Similarly, the rate of change of resistance with respect to time is given by dR/dt.

In this question, we are given dV/dt = -20.01 V/s and dR/dt = -0.03 Ω/s. We are also given V = 400 V and I = 0.08 A. We need to find how the current I is changing, which is represented by dI/dt.

To find dI/dt, we can differentiate the Ohm's law equation V = IR with respect to time:

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

Plugging in the given values, we get:

-20.01 = 400(dI/dt) + 0.08(-0.03)

Solving for dI/dt, we find that the current is changing at a rate of -0.0582 A/s.