Calculus problem (electric circuit)

[Cyannaca]

A electric circuit is made of a resistance, a solenoid and a power source.
Resistance, V= Ri
Solenoid, V = L (di/dt)
Power source, tension E(t)

Hence, L(di/dt)+ Ri = E(t). If E(t)= 20sin(5t), if L= 4 henrys and R= 20 ohms. If the intensity equals 0 at the beginning, find the amperage of i(t). Show that after a certain amount of time, the electric current becomes stationary.

I would really need help to start this problem, I'm completely lost . i was stuck at 20sin(5t)= 4(di/dt) + 20i... The answer is supposed to be
i(t)= 1/2( sin(5t)-cos(5t) +e^-5t)

 

[HallsofIvy]

A electric circuit is made of a resistance, a solenoid and a power source.
Resistance, V= Ri
Solenoid, V = L (di/dt)
Power source, tension E(t)

Hence, L(di/dt)+ Ri = E(t). If E(t)= 20sin(5t), if L= 4 henrys and R= 20 ohms. If the intensity equals 0 at the beginning, find the amperage of i(t). Show that after a certain amount of time, the electric current becomes stationary.

I would really need help to start this problem, I'm completely lost . i was stuck at 20sin(5t)= 4(di/dt) + 20i... The answer is supposed to be
i(t)= 1/2( sin(5t)-cos(5t) +e^-5t)

If 4 di/dt+ 20i= 20 sin(5t), then di/dt+ 5i= 5sin(5t). Since this is a "linear differential equation", the first thing I would do is look at the related homogeneous equation, di/dt+ 5i=0 which is the same as di/dt= -5i or
(1/i) di= -5dt. Integrating both sides, ln(i)= -5t+ C or i= C' e^-5t.

We look for a solution to the entire equation of the form
i(t)= Asin(5t)+ Bcos(5t). Then i'(t)= 5Acos(5t)- Bsin(5t). Putting those into the equation i'+ 5i= 5 sin(5t), we have 5Acos(5t)- 5Bsin(5t)+ 5Asin(5t)+5Bcos(5t)= 5sin(5t) so we must have 5A+ 5B= 0 and -5B+ 5A= 5. Adding those two equations, 10A= 5 so A= (1/2) and B= -(1/2).

Putting those together, the general solution is i(t)= C'e^-5t+ (1/2)sin(5t)- (1/2)cos(5t). i(0)= C'+ 1/2= 0 so C'= -1/2.
That is, i(t)= (1/2)[sin(5t)- cos(5t)- e^-5t].

 

[M98Ranger]

First, let's start by writing the equation that was given: L(di/dt)+ Ri = E(t). Since we are looking for the amperage, let's solve for i(t).

L(di/dt) + Ri = E(t)
L(di/dt) = E(t) - Ri
(di/dt) = (E(t) - Ri)/L

Now, we can substitute the values given in the problem:

(di/dt) = (20sin(5t) - (20)(i))/4

Next, we need to integrate both sides with respect to t to find the expression for i(t):

∫(di/dt)dt = ∫(20sin(5t) - (20)(i))/4 dt
i(t) = ∫(20sin(5t) - (20)(i))/4 dt

Let's solve the integral on the right side:

i(t) = (20/4)∫sin(5t)dt - (20/4)∫i dt
i(t) = (-4cos(5t)) - (5i/4) + C

Now, we need to find the value of C. The problem states that the intensity (i) equals 0 at the beginning, which means that i(0) = 0. Therefore, we can substitute this value in our equation:

i(0) = (-4cos(5(0))) - (5(0)/4) + C
0 = -4 + C

Thus, C = 4. We can now substitute this value back into our equation for i(t):

i(t) = (-4cos(5t)) - (5i/4) + 4

Finally, to find the stationary electric current, we need to find the value of i as t approaches infinity. As t approaches infinity, the cos(5t) term will approach 0. Therefore, the stationary electric current will be:

i = (5i/4) + 4
i = 4

This means that after a certain amount of time, the electric current will become stationary and remain at a constant value of 4 amps.