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.)
(a) What is the emf across the inductor immediately after the switch is opened?
(b) When does the current in the resistor RB have a magnitude of 1.00 mA?
www.webassign.net/katzpse1/33-p-014.png
Sorry, can not link into your textbook to see the circuits.
However just remember that the voltage V across and inductor is
V = L di/dt
and if there is NO change the inductor is just a fat wire with no resistance and in a sudden change the current through the inductor does not change instantaneously but the voltage does
To determine the instantaneous potential difference between points A and B, we need to consider the concept of inductance and its effect on the current.
(a) First, let's calculate the value of the self-induced emf (electromotive force) across the inductor immediately after the switch is opened.
The self-induced emf in an inductor is given by the formula:
emf = -L (ΔI / Δt)
Where:
emf is the self-induced emf across the inductor (unknown in this case)
L is the inductance of the circuit (given as 5.3 mH or 5.3 × 10^(-3) H)
ΔI is the change in current (given as -1.0 x 10^2 A/s, since the current is decreasing at a constant rate)
Substituting these values into the formula, we have:
emf = -(5.3 × 10^(-3) H) × (-1.0 × 10^(2) A/s)
Simplifying this expression, we get:
emf = 5.3 × 10^(-3) V/s or 5.3 mV/s
Therefore, the emf across the inductor immediately after the switch is opened is 5.3 mV/s.
(b) To find out when the current in resistor RB has a magnitude of 1.00 mA, we first need to determine the point in time at which the current reaches that value.
The potential difference across the resistor (V) can be calculated using Ohm's Law:
V = I × R
Where:
V is the potential difference across the resistor (unknown in this case)
I is the current flowing through the resistor (1.00 mA or 1.00 × 10^(-3) A)
R is the resistance of the circuit (given as 1.5 Ω)
Substituting these values into the formula, we have:
V = (1.00 × 10^(-3) A) × (1.5 Ω)
Simplifying this expression, we get:
V = 1.50 × 10^(-3) V or 1.50 mV
Therefore, the potential difference across resistor RB when the current has a magnitude of 1.00 mA is 1.50 mV.