Question

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?

Answer 1

Thanks for nothing

Answer 2

i=V*r^-1

di/dt= -V/r^2 * dr/dt

Answer 3

[AI-generated answer — Explain Bot]

To find the rate at which the current is changing, we need to differentiate the given equation I = VR with respect to time t.

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

Differentiating I = VR with respect to t using the product rule of differentiation, we have:
dI/dt = V(dR/dt) + R(dV/dt)

Since the voltage V is constant (dV/dt = 0), the equation becomes:
dI/dt = V(dR/dt)
substituting the given values:
dI/dt = 4 * (0.3)
dI/dt = 1.2 amperes per second

Therefore, the current is changing at a rate of 1.2 amperes per second.