A voltage V across a resistance R generates a current I=VR. If a constant voltage of 4 volts is put across a resistance that is increasing at a rate of 0.3 ohms per second when the resistance is 4 ohms, at what rate is the current changing? (Give units.)

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To find the rate at which the current is changing, we need to differentiate the equation I = V/R with respect to time.

Given:
V = 4 volts
R = 4 ohms
dR/dt = 0.3 ohms/s

Differentiating I = V/R with respect to time, we get:

dI/dt = (dV/dt * R - V * dR/dt) / R^2

Substituting the given values, we have:

dI/dt = (0 * 4 - 4 * 0.3) / (4^2)
= (-1.2) / 16
= -0.075 A/s

Therefore, the current is changing at a rate of -0.075 Amperes per second.

To find the rate at which the current is changing, we need to apply the chain rule of differentiation. The chain rule states that if we have a function of the form I = f(R), where f is a composite function, then the derivative of I with respect to time (dI/dt) can be calculated as (df/dR) * (dR/dt).

In this case, the function I = f(R) is given by I = VR, where V is the constant voltage. To find the derivative of I with respect to R (df/dR), we differentiate the function I = VR with respect to R and obtain df/dR = V.

Now, we need to find the rate at which the resistance is changing, given that it is increasing at a rate of 0.3 ohms per second. The rate of change of resistance (dR/dt) is given as 0.3 ohms/second.

Finally, we can calculate the rate at which the current is changing (dI/dt) by substituting the values we found:

dI/dt = (df/dR) * (dR/dt) = V * (dR/dt) = 4 volts * 0.3 ohms/second = 1.2 volts/second.

Therefore, the current is changing at a rate of 1.2 volts/second.

dR/dt = .3 ohms/s

I = 18V/R

d/dt(I) = d/dt(18V/R)

dI/dt(I) = dR/dt(18V/R)

dI/dt = dR/dt*(-18/R**2)

dI/dt = .3*(-18/4**2)

dI/dt = -.3375amp/s

Hope you understand my work! Cheerz