Question

The strength of an electrical current x flowing through the electric circuit shown is expressed as a function of time t and satisfies the following differential equation:
$\displaystyle \large{L \frac{dx}{dt} + Rx = V}$
Find the strength of the electrical current x after switch S is closed at time t = 0. Assume that L, R and V are positive constants, and also that x = 0 when t = 0. Then, find $\displaystyle \large{ \lim_{t \to \infty} x}$
Topic: Application of Differential Equation Reviews

Answer 1

Answer:

The current of the circuit at t = 0 is equal to 0.

If we take the limit as t approaches infinity, the current is equal to ε/R or V/R.

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:

$\displaystyle (cu)' = cu'$

Derivative Property [Addition/Subtraction]:

$\displaystyle (u + v)' = u' + v'$

Derivative Rule [Basic Power Rule]:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Slope Fields

• Separation of Variables

Integration

• Integrals

Integration Rule [Reverse Power Rule]:

$\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]:

$\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]:

$\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Method: U-Substitution

Electricity

Ohm's Law: V = IR

• V is voltage (in Volts)
• I is current (in Amps)
• R is resistance (in Ohms)

Circuits

• Circuit Symbols
• Kirchhoff's Laws (Loop and Junction Rule)
• Inductors

Step-by-step explanation:

*Note:

In the given equation, our variable of differentiation is x. I will rewrite this as current I for physics notation purposes.

Step 1: Define

Identify given.

$\displaystyle L \frac{dI}{dt} + RI = V$

[Assuming switch S is closed] Recall that an inductor is used in a circuit to resist change. After a long period of time, when it hits steady-state equilibrium, we expect to see the inductor act like a wire.

Step 2: Find Current Expression Pt. 1

1. [Kirchhoff's Law] Rewrite expression:
   $\displaystyle L \frac{dI}{dt} = V - IR$
2. Rewrite expression by dividing R on both sides:
   $\displaystyle \frac{L}{R} \frac{dI}{dt} = \frac{\mathcal E}{R} - I$

Step 3: Find Current Expression Pt. 2

Identify variables for u-substitution.

1. Set u:
   $\displaystyle u = \frac{\mathcal E}{R} - I$
2. [u] Differentiation [Derivative Rules and Properties]:
   $\displaystyle du = - \, dI$

Step 4: Find Current Expression Pt. 3

1. [Kirchhoff's Law] Apply U-Substitution:
   $\displaystyle - \frac{L}{R} \frac{du}{dt} = u$
2. [Kirchhoff's Law] Apply Separation of Variables:
   $\displaystyle \frac{1}{u} \, du = -\frac{L}{R} \, dt$

Recall that our initial condition is when t = 0, denoted as u₀, and we go to whatever position u we are trying to find. Also recall that time t always ranges from t = 0 (time can't be negative) and to whatever t we are trying to find.

1. [Kirchhoff's Law] Integrate both sides:
   $\displaystyle \int\limits^u_{u_0} {\frac{1}{u}} \, du = \int\limits^t_0 {- \frac{R}{L}} \, dt$
2. [Kirchhoff's Law] Rewrite [Integration Property]:
   $\displaystyle \int\limits^u_{u_0} {\frac{1}{u}} \, du = - \frac{R}{L} \int\limits^t_0 {} \, dt$
3. [1st Integral] Apply Logarithmic Integration:
   $\displaystyle \ln | u | \bigg| \limits^u_{u_0} = - \frac{R}{L} \int\limits^t_0 {} \, dt$
4. [2nd Integral] Apply Integration Rule [Reverse Power Rule]:
   $\displaystyle \ln | u | \bigg| \limits^u_{u_0} = - \frac{R}{L} t \bigg| \limits^t_0$
5. Apply Integration Rule [Fundamental Theorem of Calculus 1]:
   $\displaystyle \ln | \frac{u}{u_0} | = - \frac{R}{L} t$
6. Apply e to both sides:
   $\displaystyle e^{\ln | \frac{u}{u_0} |} = e^{- \frac{R}{L} t}$
7. Simplify:
   $\displaystyle \frac{u}{u_0} = e^{- \frac{R}{L} t}$
8. Rewrite:
   $\displaystyle u = u_0 e^{- \frac{R}{L} t}$

Recall that our initial condition u₀ (derived from Ohm's Law) contains only the voltage across resistor R, where voltage is supplied by the given battery. This is because the current is stopped once it reaches the inductor in the circuit since it resists change.

1. Back-Substitute in u and u₀:
   $\displaystyle \frac{\mathcal E}{R} - I = \frac{\mathcal E}{R} e^{- \frac{R}{L} t}$
2. Solve for I:
   $\displaystyle I = \frac{\mathcal E}{R} - \frac{\mathcal E}{R} e^{- \frac{R}{L}t}$

Step 5: Solve

If we are trying to find the strength of the electrical current I at t = 0, we simply substitute t = 0 into our current function:

$\displaystyle\begin{aligned}I(t) & = \frac{\mathcal E}{R} - \frac{\mathcal E}{R} e^{- \frac{R}{L}t} \\I(0) & = \frac{\mathcal E}{R} - \frac{\mathcal E}{R} e^{- \frac{R}{L}(0)} \\& = \boxed{\bold{0}}\end{aligned}$

If we are taking the limit as t approaches infinity of the current function I(t), we are simply just trying to find the current after a long period of time, which then would just be steady-state equilibrium:

$\displaystyle\begin{aligned}I(t) & = \frac{\mathcal E}{R} - \frac{\mathcal E}{R} e^{- \frac{R}{L}t} \\\lim_{t \to \infty} I(t) & = \frac{\mathcal E}{R} - \frac{\mathcal E}{R} e^{- \frac{R}{L}(\infty)} \\& = \boxed{\bold{\frac{\mathcal E}{R}}}\end{aligned}$

∴ we have found the current I at t = 0 and the current I after a long period of time and proved that an inductor resists current running through it in the beginning and acts like a wire when in electrical equilibrium.

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Topic: AP Physics C - EMAG

Unit: Induction