# I RL circuit with polynomial forcing function solution (units)

#### K Murty

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.

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#### Delta2

Homework Helper
Gold Member
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 $s^2$).

#### K Murty

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.

#### K Murty

i will assume $$t \,\, (V)$$ and solve it again for the source term as $$\dfrac{t}{L} \,\, \frac{V}{H}$$

#### Delta2

Homework Helper
Gold Member
Yes the initial DE has to be $iR+L\frac{di}{dt}=V_0t$ where $V_0=1\frac{V}{s}$

#### K Murty

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.

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#### Delta2

Homework Helper
Gold Member
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.

"RL circuit with polynomial forcing function solution (units)"

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