Answer:
= - 1.3 x 10⁻⁴ A/s
Explanation:
We take the derivative of Ohm's law with respect to time
V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.02 V/s,
resistance is increasing at 0.04 ohm/s,
resistance itself is 300 ohms,
and current is 0.02 A
Substituting:
-0.02 V/s = (0.02 A)(0.04 ohm/s) + (300 ohms)(dI/dt)
dI/dt = -1.3 x 10⁻⁴ A/s
Given Information:
Rate of change of Voltage = dV/dt = -0.02 V/s
Rate of change of Resistance = dR/dt = 0.04 Ω /s
Resistance = R = 300 Ω
Current = I = 0.02 A
Required Information:
Rate of change of current = dI/dt = ?
Answer:
dI/dt = -6.93x10⁻⁵ A/s
Explanation:
The Ohm's law is given by
V = IR eq. 1
Where V is the voltage across Resistance R and I is the current flowing through the Resistance R
Taking derivative of eq. 1 with respect to time t yields
dV/dt = I*dR/dt + dI/dt*R
dI/dt*R = -I*dR/dt + dV/dt
dI/dt = (-I*dR/dt + dV/dt)/R eq. 2
Now substitute the given values into eq. 2
dI/dt = (-0.02*0.04 + (-0.02))/300
dI/dt = (-0.0008 - 0.02)/300
dI/dt = -0.0208/300
dI/dt = -6.93x10⁻⁵ A/s
Therefore, the current is decreasing at the rate of 6.93x10⁻⁵ A/s
