I RL circuit with polynomial forcing function solution (units)

  • Thread starter K Murty
  • Start date
34
17
Hello.
This is the differential equation.

$$
i \cdot \dfrac{R}{L} + \dfrac{di}{dt} = t
$$
My solution path:
Homogenous solution:
$$
r + \dfrac{R}{L} = 0 \\ \\

r = -\dfrac{R}{L} \\ \\

i_{h}(t) = C_{0} e^{ -\dfrac{R}{L} t} \\ \\ $$

Particular solution:

Try $$ y_{p} = at + b \\ \\

y_{p} \dfrac{R }{L } + y_{p}' = t \\ \\

(at + b) \dfrac{R}{L} + a = t \\ \\

at\dfrac{R}{L} = t \\ \\

b \dfrac{R}{L} + a = 0 \\ \\

a = \dfrac{L}{R} \\ \\

b = - \dfrac{ L^2}{R^2} \\ \\ $$

Verifying particular solution:

$$ y_{p} = \dfrac{L}{R} \cdot ( t - \dfrac{L}{R} ) \\ \\

y_{p}' = \dfrac{L}{R} \\ \\

y_{p} \dfrac{R}{L} + y_{p}= t \\ \\

\dfrac{L}{R} \cdot ( t - \dfrac{L}{R} ) \cdot \dfrac{R}{L} + \dfrac{L}{R} = t \\ \\

t - \dfrac{L}{R} + \dfrac{L}{R} = t $$
Boundary conditions given:
$$
\boxed{I(0) = 0 } \text{inital conditions} \\ \\

0 = C_{0} - \dfrac{L^2}{R^2} \\ \\

C_{0} = \dfrac{L^2}{R^2} \\ \\

I(t) = \dfrac{L^2}{R^2} e^{ -\frac{R}{L} t} + \dfrac{ L}{ R} ( t - \dfrac{ L}{ R}) \\ \\

\boxed{i(t) = \dfrac{L^2}{R^2} \cdot e^{( -\frac{R}{L} t)} + \left( \dfrac{ L}{ R}( t - \dfrac{ L}{ R}) \right) \,\,\,\, (\text{A} ) } \,\,\,\, \text{Complete solution} $$
My question is, despite the complete solution fitting the boundary conditions, the units must be amperes, and I dont think the units are amperes. How can i get a solution with the correct units? My solution path seems correct to me as t is a first order polynomial.
 
Last edited:

Delta2

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The problem with the units is right from the start of the ODE

##I\frac{R}{L}+\frac{dI}{dt}=t## (1)

because the left hand side of (1) has units of ##\frac{A}{s}## while the right hand side has units of just ##s## (seconds). So to fix this you must multiply only the right hand side by a constant C which will be just 1 but will have units of ##\frac{A}{s^2}## so ##C=1\frac{A}{s^2}## so the differential equation will be

##I\frac{R}{L}+\frac{dI}{dt}=Ct##(2) and now the units in both sides of (2) are ##\frac{A}{s}##.

(or alternatively you can multiply only the left hand side of (1) by a constant D such that ##D=1\frac{s^2}{A}##.

Your final complete solution will not change at all as number, because it will be just multiplied by C which is just 1, but it will change as units and it will be amperes. (##\frac{A}{s^2}s^2=A## now as it is, the units are ##s^2##).
 
34
17
The problem with the units is right from the start of the ODE

##I\frac{R}{L}+\frac{dI}{dt}=t## (1)

because the left hand side of (1) has units of ##\frac{A}{s}## while the right hand side has units of just ##s## (seconds). So to fix this you must multiply only the right hand side by a constant C which will be just 1 but will have units of ##\frac{A}{s^2}## so ##C=1\frac{A}{s^2}## so the differential equation will be

##I\frac{R}{L}+\frac{dI}{dt}=Ct##(2) and now the units in both sides of (2) are ##\frac{A}{s}##.

(or alternatively you can multiply only the left hand side of (1) by a constant D such that ##D=1\frac{s^2}{A}##.

Your final complete solution will not change at all as number, because it will be just multiplied by C which is just 1, but it will change as units and it will be amperes. (##\frac{A}{s^2}s^2=A## now as it is, the units are ##s^2##).
I made a mistake in my solution.
The initial DE ACTUALLY given was:
$$
iR + L\dfrac{di}{dt} = t
$$
Multiplying the DE by $$ \dfrac{1}{L}$$
Gives:
$$ i \dfrac{R}{L} + \dfrac{di}{dt} = \dfrac{t}{L} $$
I hurried the solution and forgot to multiply the source term by $$ \dfrac{1}{L} $$
But even without multiplying the entire equation, the initial units are:
$$ V = s $$
Which is wrong. So I think this question means that there is a voltage source which is outputs the voltage equal to the duration of time.
My apologies, the person who gave it to me to solve was not clear about it.
 
34
17
i will assume $$ t \,\, (V) $$ and solve it again for the source term as $$ \dfrac{t}{L} \,\, \frac{V}{H} $$
 

Delta2

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Yes the initial DE has to be ##iR+L\frac{di}{dt}=V_0t## where ##V_0=1\frac{V}{s}##
 
34
17
Yes the initial DE has to be ##iR+L\frac{di}{dt}=V_0t## where ##V_0=1\frac{V}{s}##
Yes. Here is the solution that was realised through your insight:
$$
V_{0} = \frac{{V} }{ s} \\ \\
i\dfrac{R}{L} + \dfrac{di}{dt} = \dfrac{ V_{0} t}{L} \\
\dfrac{R}{L} + r = 0 \\
r = - \dfrac{R}{L} $$

$$ i_{h}(t) = C_{0} e^{(- \dfrac{R}{L} t )} \\

\text{Particular solution:}$$

$$ i_{p}(t) = at + b $$

$$ (at + b ) \dfrac{R}{L} + a = \dfrac{V_{0}t}{L} $$

$$ at \frac{R}{L} + (a + b) = \dfrac{V_{0}t}{L} $$

$$ at \frac{R}{L} = \dfrac{V_{0}t}{L} $$

$$ \boxed{ a = \dfrac{ V_{0} }{ {R} } } \tag{a}$$

$$ \dfrac{V_{0} }{ {R} } + b \frac{R}{L} = 0 $$

$$ \boxed{ b = - \dfrac{ V_{0} L}{ {R^2} } } \tag{b} $$

$$ i(t) = C_{0} e^{(- \frac{R}{L} t )} + (\dfrac{ V_{0} }{ {R} } t + - \dfrac{ V_{0} L}{ {R^2} } ) \tag{} $$

$$ \text{Initial values to solve for constants}$$

$$ i(0) = 0 \\ \\

0 = C_{0} - \dfrac{ V_{0} L}{ {R^2} } \\ \\

C_{0} = \dfrac{ V_{0} L}{ {R^2} } \\ \\

{ i(t) = \frac{ V_{0} L }{ {R^2} }e^{(- \frac{R}{L} t )} +\dfrac{ V_{0} }{ R } t - \dfrac{ V_{0} L}{ {R^2} } } \\ \\ $$

$$ \boxed{ i(t) = \dfrac{ V_{0} L}{ {R^2} } (e^{(- \frac{R}{L} t )} - 1 ) +\dfrac{ V_{0} }{ R } t } \tag{Complete}
$$
The units work out now:
$$ \frac{ \frac{V}{s} \cdot H }{\Omega^2} + \frac{ \frac{V}{s} \cdot s}{\Omega} = A \\
\frac{V F }{ s} + \frac{V}{\Omega} = A \\
(\frac{ V}{ s} \frac{Q}{V} = \frac{Q}{s}=A) + A = A
$$
Thank you.
 
Last edited:

Delta2

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I just checked the units of the complete solution,the first term is ##\frac{V}{s Ohm}\frac{H}{Ohm}=\frac{A}{s}s=A## and the second term
##\frac{V}{s}\frac{s}{Ohm}=\frac{V}{Ohm}=A## so everything looks fine.
 

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