Thus, { which reduces to . Assuming U = 0 at t = 0 and integrating I Where U is the potential energy of the inductor at any time t and is the current at that instant. Example VIII–8 In the circuit sketched at the an inductor having inductance L and E. resistance r is connected in parallel with a resistance R and a battery having an EMF, After switch S has been closed a long time it is suddenly opened. E, Find the following currents in terms of L, R, and r. I through the inductor before the switch is opened. Solution: Since the current anywhere in the circuit is steady and the circuit acts as if L were not present except for its resistance r. The circuit then is E. effectively and thus E Ir, E/r. Therefore, since = the current I through L immediately after S is opened. Io I E/r Solution: Since and = thru r just before S is opened = and t = 0 E/r = (E/r)= I through R before S is opened. E IR. IR E/R. Solution: The situation is the same as in (i) and VR = = Therefore, = I through R at t = after S is opened. Solution: Since r, R, and L are all in series (E/r) = 0.37(E/r) Find immediately after S is opened. E/r E. Solution: at t = 0. = (see a - ii) and thus Write the differential equation for the current and derive the expression for the current in L I as a function of time after the switch is opened. Sketch the graph of vs t. Solution: Applying K-2 to the circuit after the switch is opened:
Thus, { which reduces to . Assuming U = 0 at t = 0 and integrating Where U is the potential energy of the inductor at any time t and I is the current at that instant. Example VIII–8 In the circuit sketched at the an inductor having inductance L and resistance r is connected in parallel with a resistance R and a battery having an EMF, E. After switch S has been closed a long time it is suddenly opened. Find the following currents in terms of E, L, R, and r. I through the inductor before the switch is opened. Solution: Since the current anywhere in the circuit is steady and the circuit acts as if L were not present except for its resistance r. The circuit then is effectively and thus E. Therefore, since E = Ir, the current E/r. I through L immediately after S is opened. Solution: Since and Io = I thru r just before S is opened = E/r and t = 0 = (E/r)= E/r I through R before S is opened. Solution: The situation is the same as in (i) and VR = E = IR. Therefore, IR = E/R. I through R at t = after S is opened. Solution: Since r, R, and L are all in series (E/r) = 0.37(E/r) Find immediately after S is opened. Solution: at t = 0. = E/r (see a - ii) and thus E. Write the differential equation for the current and derive the expression for the current in L as a function of time after the switch is opened. Sketch the graph of I vs t. Solution: Applying K-2 to the circuit after the switch is opened: