Series Circuit DIfferential Equation - My answer is coming out to be wrong

In summary: So, in summary, the correct equation to use for this problem is L di/dt + Ri = E(t), and the electromotive force needed to achieve a current of 6.5 amps in 1.25 seconds, with a steady current of 2 amps already present, is 15.1 volts.
  • #1
Arshad_Physic
51
1
Series Circuit DIfferential Equation - My answer is coming out to be wrong...

Homework Statement



A 6.5 amp current is necessary in order to start the motor for a certain factory appliance. The circuit for the appliance is designed with a 2 ohm resistor and a 1.5 henry inductor. If there already exists a steady current of 2 amps in the circuit at time t = 0, what electromotive force is necessary to achieve the desired current in 1.25 seconds?

The answer should be 15.1 volts.

Homework Equations



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

or

2) R dq/dt + q/C = E(t)

The Attempt at a Solution



It makes sense to me to use Equation (1):

L di/dt + Ri = E(t)

1.5 di/dt + 2i = E(t)

=> di/dt = [E(t) - 2i ] /1.5

=> di/[E(t) - 2i] = dt/1.5

=> ln[E(t) - 2i) = t/1.5 + c

=> E(t) = Ae-t/1.5 + 2i

i(o) = 2

A = E - 4

=> E(t) = (E-4)e-t/1.5 + 2i

i(1.25) = 6.5

=> E(t) = (E-4)e-1.25/1.5 + 13

=> E - (E-4)e-0.83333 = 13

=> Ee0.83333 - (E-4) = 13e-0.83333

=> Ee-0.83333 - E + 4 = 29.9126

=> E(e-0.83333 - 1) = 25.9126

=> E = 19.9178 Ans.

The answer should be 15.1 volts. PLease help! What is the thing that I am doing wrong!

Thanks!
 
Physics news on Phys.org
  • #2


First off, Your entire method could not possibly work if E(t) is not simply a constant, but thankfully in this case it is. You only made one simple mistake when you integrated di/[E-2i]. you're missing a -1/2 in front of the ln|E-2i|.
 

What is a series circuit differential equation?

A series circuit differential equation is a mathematical representation of a series circuit, which is a circuit in which the components are connected in a single loop. The equation describes the relationship between the voltage, current, and resistance in the circuit.

Why is my answer for the series circuit differential equation coming out to be wrong?

There could be several reasons for an incorrect answer to the series circuit differential equation. Some possible causes could be errors in calculations, incorrect values for the circuit components, or incorrect application of the differential equation formula.

What are the key components of a series circuit differential equation?

The key components of a series circuit differential equation are the voltage (V), current (I), and resistance (R) in the circuit. These variables are related by the equation V = IR, where V is the voltage, I is the current, and R is the resistance.

How do I solve a series circuit differential equation?

To solve a series circuit differential equation, you can use the standard formula V = IR and substitute in the values for the voltage, current, and resistance in the circuit. You can then rearrange the equation to solve for any of the three variables.

What are some real-world applications of series circuit differential equations?

Series circuit differential equations have many real-world applications, including in electrical engineering, physics, and electronics. They are used to analyze and predict the behavior of series circuits, such as in calculating the voltage drop across different components or determining the total resistance of a circuit.

Similar threads

  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
967
  • Engineering and Comp Sci Homework Help
Replies
28
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
Back
Top