This set of Ordinary Differential Equations Quiz focuses on “Simple Electrical Networks Solution”.
1. A constant electromotive force E volts is applied to a circuit containing a constant resistance R ohm in series with a constant inductance L henries. If the initial current is zero. What is the current in the circuit at any time t?
a) \(\frac{E}{R} \left(1-e^{\frac{-Rt}{L}}\right)\)
b) \(\frac{E}{R} \left(e^{\frac{Rt}{L}}\right)\)
c) \(\frac{E}{L} (1-e^{RLt})\)
d) \(\frac{E}{L} (e^{-RLt})\)
View Answer
Explanation: The differential equation for the circuit is derived based on Kirchhoff’s voltage law and it is given by \(L \frac{di}{dt} + Ri = E \rightarrow \frac{di}{dt} + \frac{R}{L} \,i = \frac{E}{L}\) this is of the form \(\frac{dy}{dx} + Px = Q\)
i.e it is linear DE \( I.F = e^{\int P \,dx} = e^{\int \frac{R}{L} \,dt} = e^{\frac{Rt}{L}}, Q=\frac{E}{L} \) and its solution is given by
\(ye^{\int P \,dx} = \int Q e^{\int P \,dx} \,dx + c \rightarrow \,ie^{\frac{Rt}{L}} = \int e^{\frac{Rt}{L}} * \frac{E}{L} \,dt + c\)
\(ie^{\frac{Rt}{L}} = (e^{\frac{Rt}{L}} * \frac{E}{L} * \frac{1}{R/L}) + c \rightarrow i = \frac{E}{R} + ce^{\frac{-Rt}{L}}\)….(1) given i(0)=0
\(\rightarrow -\frac{E}{R} = c\) substituting in (1) we get \(i(t) = \frac{E}{R}-\frac{E}{R} e^{\frac{-Rt}{L}} = \frac{E}{R} \left(1-e^{\frac{-Rt}{L}}\right).\)
2. A voltage Ee-at is applied at t=0 to a circuit containing inductance L and resistance R. Determine the current at any time t.
a) \(\frac{Ee^{-at}}{L-Ra} \left(e^{\frac{-Rt}{L}}\right)\)
b) \(\frac{E}{R-La} \left(e^{-at}-e^{\frac{-Rt}{L}}\right)\)
c) \(\frac{E}{La} \left(e^{-at}+e^{\frac{Rt}{L}}\right)\)
d) \(\frac{Ee^{-at}}{R} \left(1-e^{\frac{-Rt}{L}}\right)\)
View Answer
Explanation: The differential equation for the circuit is derived based on Kirchhoff’s voltage law and it is given by \(L \frac{di}{dt} + Ri = E \rightarrow \frac{di}{dt} + \frac{R}{L} \,i = \frac{Ee^{-at}}{L}\) this is of the form \(\frac{dy}{dx} + Px = Q\)
its solution is \(ie^{\frac{Rt}{L}} = \int e^{\frac{Rt}{L}} * \frac{Ee^{-at}}{L} \,dt + c = \frac{E}{L} \int e^{(\frac{R}{L}-a)t} \,dt + c\)
\(ie^{\frac{Rt}{L}} = \Big\{\frac{E}{L} * e^{(\frac{R}{L}-a)t} * \frac{1}{(\frac{R}{L}-a)}\Big\} + c = \frac{E}{R-La} * e^{(\frac{R}{L}-a)t} + c\)
\(i = \frac{Ee^{-at}}{L-Ra} + ce^{\frac{-Rt}{L}}\)….(1) using i(0)=0 we get \(c=\frac{-E}{R-La} \)
substituting in (1)
\(i(t) = \frac{E}{R-La} \left(e^{-at}-e^{\frac{-Rt}{L}}\right)\).
3. 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) \(\frac{10}{R^2+L^2} (R cost – L sint + Le^{\frac{Rt}{L}})\)
b) \(\frac{10}{R^2+L^2} (R sint + L cost – Le^{\frac{Rt}{L}})\)
c) \(\frac{10}{R^2+L^2} (R sint – L cost + Le^{\frac{-Rt}{L}})\)
d) \(\frac{10}{R^2+L^2} (L cost – R sint + Le^{\frac{-Rt}{L}})\)
View Answer
Explanation: \(L \frac{di}{dt} + Ri = 10 \,sint \rightarrow \frac{di}{dt} + \frac{R}{L} i=\frac{10 sin t}{L} \) since it is an Linear DE its solution is
given by \(ie^{\frac{Rt}{L}} = \int \frac{10}{L} * sint * e^{\frac{Rt}{L}} \,dt = \frac{10}{L} \int \,sint * e^{\frac{Rt}{L}} \,dt \) ……using the formula
\(\int e^{at} \,sin\, bt \,dt = \frac{e^{at}}{a^2+b^2} (a sinbt – b cosbt)\) we get
\(ie^{\frac{Rt}{L}} = \frac{10}{L} * \frac{e^{\frac{Rt}{L}}}{(\frac{R}{L})^2+1^2} (\frac{R}{L} sint-cost)+c\)
\(ie^{\frac{Rt}{L}} = \frac{10e^{\frac{Rt}{L}}} {R^2+L^2} (R sin t-L cos t) + c \rightarrow i(t) = \frac{10}{R^2+L^2} (R sint-L cost) + ce^{\frac{-Rt}{L}}\)..(1)
using i(0)=0 we have \(c=\frac{-10L}{R^2+L^2} \) substituting this back in (1) we get
\(i(t) = \frac{10}{R^2+L^2} (R sint-L cost) – \frac{10L}{R^2+L^2} e^{\frac{-Rt}{L}} = \frac{10}{R^2+L^2} (R sint – L cost + Le^{\frac{-Rt}{L}}).\)
Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.
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