A voltage V across a resistance R generates a current I=V/R. If a constant voltage of 16 volts is put across a resistance that is increasing at a rate of 0.2 ohms per second when the resistance is 6 ohms, at what rate is the current changing? (Give units.)

-0.05

To find the rate at which the current is changing, we can use the chain rule of differentiation in calculus. The formula for the current through a resistance is I = V/R, where V is the voltage and R is the resistance. We are given that V = 16 volts and dR/dt = 0.2 ohms per second. We need to find dI/dt, the rate at which the current is changing.

To do this, we'll differentiate the formula I = V/R with respect to time t:

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

Since V is constant at 16 volts, dV/dt is zero, and the formula simplifies to:

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

Applying the given values, we have:

dI/dt = (-16 volts * 0.2 ohms per second) / (6 ohms)^2

Calculating this, we get:

dI/dt = -0.1778 Amperes per second

So, the current is changing at a rate of approximately -0.1778 Amperes per second. The negative sign indicates that the current is decreasing.

-4/45 amp/sec

I agree