At one instant, a current of 8.0 A flows through part of a circuit as shown in the figure below. Determine the instantaneous potential difference between points A and B if the current starts to decrease at a constant rate of 1.0 ✕ 102 A/s. (Assume that R = 1.5 Ω and L = 5.3 mH.
Assume current is flowing from point A to point B
webassign.net/katzpse1/33-p-012-alt.png
To determine the instantaneous potential difference between points A and B, we need to consider two factors: the voltage across the resistor (VR) and the voltage across the inductor (VL).
To calculate the voltage across the resistor, we can use Ohm's Law: VR = IR, where IR is the current flowing through the resistor and R is the resistance.
In this case, the current flowing through the resistor is given as 8.0 A, and the resistance is given as 1.5 Ω. Therefore, VR = 8.0 A * 1.5 Ω = 12 V.
Next, we need to calculate the voltage across the inductor. The voltage across an inductor is given by the formula VL = L * dI/dt, where L is the inductance and dI/dt is the rate at which the current is changing.
In this case, the inductance is given as 5.3 mH, which is equivalent to 5.3 * 10^-3 H. The rate at which the current is changing is given as 1.0 * 10^2 A/s. Therefore, VL = (5.3 * 10^-3 H) * (1.0 * 10^2 A/s) = 0.53 V.
To determine the total potential difference between points A and B, we can sum up the voltage across the resistor and the voltage across the inductor: VAB = VR + VL = 12 V + 0.53 V = 12.53 V.
Therefore, the instantaneous potential difference between points A and B is 12.53 V.