After being closed for a long time, the switch S in the circuit shown in the figure below is thrown open at
t = 0. In the circuit, emf = 24.0 V, RA = 4.10 kΩ, RB = 6.80 kΩ, and L = 608 mH.
(a) What is the emf across the inductor immediately after the switch is opened?
(b) What is the common time constant for the two circuits?
www.webassign.net/katzpse1/33-p-014.png
To find the emf across the inductor immediately after the switch is opened, we can use the formula for the voltage across an inductor in an RL circuit:
V = L(di/dt)
where V is the voltage across the inductor, L is the inductance, and (di/dt) is the rate of change of current with respect to time.
(a) In this case, the switch is opened, so the current through the inductor will decrease over time. However, immediately after the switch is opened, the current will be at its maximum value. This means that the rate of change of current with respect to time (di/dt) is initially equal to zero.
Therefore, immediately after the switch is opened, the emf across the inductor is also equal to zero. So the answer to part (a) is 0 V.
(b) The common time constant for the two circuits can be found using the formula:
τ = (RA || RB) * L
where τ (tau) is the time constant, RA || RB represents the total resistance in the circuit, and L is the inductance.
To find the total resistance in the circuit (RA || RB), we use the formula for resistors in parallel:
1 / (RA || RB) = 1 / RA + 1 / RB
Substituting the given values of RA = 4.10 kΩ and RB = 6.80 kΩ into the formula, we can find the value of RA || RB.
Once we have the value of RA || RB, we can substitute it along with the given value of L = 608 mH into the formula to calculate the time constant τ.
Note: To calculate the value of RA || RB, the resistances RA and RB should be converted to ohms (multiply them by 1000), and the inductance L should be converted to henries (divide it by 1000).
I hope this helps you understand how to find the answers to the given questions.