250+ TOP MCQs on Simple Electrical Networks Solution and Answers

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}))
Answer: a
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))
Answer: b
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 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}}))
Answer: c
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}}).)

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