A voltage V across a resistance R generates a current I=V/R. If a constant voltage of 17 volts is put across a resistance that is increasing at a rate of 0.3 ohms per second when the resistance is 7 ohms, at what rate is the current changing? (Give units.)

I = 17/R

dI/dt = -17/R^2 dR/dt
now plug in your numbers.
dI/dt = -17/7^2 * 0.3 = -0.104 amps/sec

Well, well, well, looks like we have a classic case of electrical fun! Let's get to the bottom of it.

We know that Ohm's Law tells us that the current, I, passing through a resistor is equal to the voltage, V, divided by the resistance, R. So, I = V/R.

To find out the rate at which the current is changing, we need to take the derivative of this equation with respect to time. Since we have the rate of change of the resistance, we can just plug in the values!

Differentiating both sides of the equation, we get:

dI/dt = (dV/dt * R - V * dR/dt) / R²

Now, let's plug in the values given! We have a constant voltage of 17 volts, and the resistance is increasing at a rate of 0.3 ohms per second when it's at 7 ohms. Math time!

dI/dt = (0 * 7 - 17 * 0.3) / 7²

Simplifying this gives us:

dI/dt = -0.3 / 49

So, the rate at which the current is changing is approximately -0.0061 A/s, where A/s stands for amperes per second.

And there you have it! Just a little jolt of humor to help you through the electrifying journey.

To find the rate at which the current is changing, we need to differentiate the current equation with respect to time.

Given:
Voltage V = 17 volts
Resistance R = 7 ohms
Rate of change of resistance dR/dt = 0.3 ohms per second

Current I = V/R

Differentiating both sides of the equation with respect to time (t):

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

Since the voltage is constant (dV/dt = 0), we can simplify the equation:

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

Substituting the given values:

dI/dt = - (17 volts * 0.3 ohms per second) / (7 ohms)^2

Calculating:

dI/dt = - (5.1 volts per second) / 49 ohms

Therefore, the rate at which the current is changing is approximately -0.104 amps per second.

To find the rate at which the current is changing, we need to find the derivative of the current with respect to time.

Given:
Voltage, V = 17 volts
Resistance, R = increasing at a rate of 0.3 ohms per second
Initial resistance, R₀ = 7 ohms

We know that the current, I, is given by I = V / R. Therefore, we need to find the derivative of I with respect to time, which can be written as dI/dt.

Let's find the derivative of I with respect to time, considering R as a function of time:

I = V / R(t)

Taking the derivative of both sides with respect to time:

dI/dt = d/dt(V / R(t))

To differentiate V / R(t), we can use the quotient rule:

dI/dt = (dV/dt * R(t) - V * dR(t)/dt) / [R(t)]²

Now, let's plug in the given values:
dV/dt = 0, since the voltage is constant
dR(t)/dt = 0.3 ohms per second
R(t) = R₀ + dR(t)/dt * t = 7 + 0.3t

Substituting these values into the derivative equation:

dI/dt = (0 * [7 + 0.3t] - 17 * 0.3) / ([7 + 0.3t])²

Simplifying the equation:

dI/dt = -5.1 / ([7 + 0.3t])²

Therefore, the rate of change of the current is -5.1 / ([7 + 0.3t])² amps per second.