Question

The current i amperes at any time t is given by \(L \frac{di}{dt} + Ri = E\), when a resistance R ohms is connected in series with an inductance L henries and E.M.F of E volts. If E=10 sin(t) volts and i=0 when t=0,which among the following is the correct expression for i(t) at any time t?

(a) \(\frac{10}{R^2+L^2} (R cos⁡t – L sin⁡t + Le^{\frac{Rt}{L}})\)

(b) \(\frac{10}{R^2+L^2} (R sin⁡t + L cos⁡t – Le^{\frac{Rt}{L}})\)

(c) \(\frac{10}{R^2+L^2} (R sin⁡t – L cos⁡t + Le^{\frac{-Rt}{L}})\)

(d) \(\frac{10}{R^2+L^2} (L cos⁡t – R sin⁡t + Le^{\frac{-Rt}{L}})\)

The question was posed to me in an online quiz.

My doubt is from Simple Electrical Networks Solution topic in section Ordinary Differential Equations – First Order & First Degree of Engineering Mathematics

Answer 1

Right answer is (c) \(\frac{10}{R^2+L^2} (R sin⁡t – L cos⁡t + Le^{\frac{-Rt}{L}})\)

Easy explanation: \(L \frac{di}{dt} + Ri = 10 \,sin⁡t \rightarrow \frac{di}{dt} + \frac{R}{L} i=\frac{10 sin t}{L} \) since it is an Linear DE its solution is

given by \(ie^{\frac{Rt}{L}} = \int \frac{10}{L} * sin⁡t * e^{\frac{Rt}{L}} \,dt = \frac{10}{L} \int \,sin⁡t * e^{\frac{Rt}{L}}  \,dt \)  ……using the formula

\(\int e^{at}  \,sin\, ⁡bt \,dt = \frac{e^{at}}{a^2+b^2}  (a sin⁡bt – b cos⁡bt)\)  we get

\(ie^{\frac{Rt}{L}} = \frac{10}{L} * \frac{e^{\frac{Rt}{L}}}{(\frac{R}{L})^2+1^2} (\frac{R}{L} sin⁡t-cos⁡t)+c\)

\(ie^{\frac{Rt}{L}} = \frac{10e^{\frac{Rt}{L}}} {R^2+L^2} (R sin t-L cos t) + c \rightarrow i(t) =  \frac{10}{R^2+L^2} (R sin⁡t-L cos⁡t) + ce^{\frac{-Rt}{L}}\)..(1)

using i(0)=0 we have \(c=\frac{-10L}{R^2+L^2} \) substituting this back in (1) we get

\(i(t) =  \frac{10}{R^2+L^2} (R sin⁡t-L cos⁡t) – \frac{10L}{R^2+L^2} e^{\frac{-Rt}{L}} = \frac{10}{R^2+L^2} (R sin⁡t – L cos⁡t + Le^{\frac{-Rt}{L}}).\)