A motor is designed to operate on 117 V and draws a current of 12.3 A when it first starts up. At its normal operating speed, the motor draws a current of 3.38 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed

Now that we have a feel for the state of the circuit in its steady state, let us obtain the expression for the current in the circuit as a function of time. note that we can use the loop rule (going around counterclockwise): e−vr−vl=0. note as well that vr=ir and vl=ldidt. using these equations, we can get, after some rearranging of the variables and making the subsitution x=er−i, dxx=−rldt. integrating both sides of this equation yields x=x0e−rt/l. use this last expression to obtain an expression for i(t). remember that x=er−i and that i0=i(0)=0. express your answer in terms of e, r, and l. you may or may not need all these variables. use the notation exp(x) for ex.