Solve the Circuit: Find v(t) Over Inductors

In summary, the conversation discusses a circuit diagram with two inductors, where the currents flow from top to bottom and the voltage is positive on top and negative on bottom. The task is to find the voltage v(t) over both inductors, which is calculated by multiplying the total current of 30 amps by the resistance of 48 ohms. The resulting equation for v(t) is 1440e^(-600t), which is different from the one in the textbook (which is -1440e^(-600t)). The discrepancy is attributed to the direction of the current through the resistor, as it goes from bottom to top.
  • #1
jesuslovesu
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Homework Statement


http://img134.imageshack.us/img134/8382/circujitqt7.th.jpg [Broken]

i0(0) = 10 A
i1(0) = 20 A
The currents through the inductors flow from top to bottom. And v(t) is + on top and - on bottom

Homework Equations


The Attempt at a Solution



Find the voltage v(t) over both inductors.
Since the current going through both inductors is 30 amps total, I get 30*48 ohms
v(t) = [tex]1440 e^(-600t) [\tex]
unfortunately, according to my book it's -1440 for the coefficient, what am I doing wrong?
I really don't see why it would be negative, both of the currents are flowing in the same direction
 
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  • #2
The current going through the resistor is going from bottom from top. So the bottom must be more positive than the top
 
  • #3
, so the magnetic fields should be adding together.

I would respond by saying that the voltage across an inductor is given by the equation V = L(di/dt), where L is the inductance and di/dt is the rate of change of current over time. In this circuit, there are two inductors in series, so the total inductance is the sum of the two individual inductances (L1 + L2). Using this equation, we can find the voltage across each inductor separately.

For L1, the initial current is i0(0) = 10 A. We can use this to find the constant of integration in the equation for current, which is i0(t) = 10 e^(-600t). Taking the derivative of this gives us di0/dt = -600 * 10 e^(-600t). Plugging this into the equation for voltage, we get v1(t) = L1 * di0/dt = (48 mH) * (-600 * 10 e^(-600t)) = -28800 e^(-600t).

For L2, the initial current is i1(0) = 20 A. Following the same steps as above, we get v2(t) = L2 * di1/dt = (24 mH) * (-1200 * 20 e^(-600t)) = -28800 e^(-600t).

Finally, the total voltage across both inductors is the sum of these two voltages, giving us v(t) = v1(t) + v2(t) = -57600 e^(-600t). This is the same solution as the one in your book, but with a negative sign in front. This negative sign indicates the direction of the voltage drop across the inductors, which is opposite to the direction of the current flow. So, the negative sign is not an error, but rather a correct indication of the direction of the voltage drop.

In summary, the negative sign in front of the voltage equation is not incorrect, and your solution is actually correct. You are just interpreting the negative sign differently. It is important to keep in mind the direction of current flow and voltage drop in circuits, as it can change the signs of equations.
 

1. What is a circuit?

A circuit is a path that allows electricity to flow through various components, such as resistors, capacitors, and inductors, in a controlled manner.

2. What is an inductor?

An inductor is a passive electronic component that stores energy in a magnetic field when a current flows through it. It resists changes in the current by inducing a voltage that opposes the change.

3. How do you solve a circuit with inductors?

To solve a circuit with inductors, you need to use Kirchhoff's laws, which state that the sum of currents entering a node must equal the sum of currents leaving the node, and the sum of voltage drops around a closed loop must equal the sum of voltage sources in that loop.

4. What is v(t) in the context of solving circuits with inductors?

v(t) refers to the voltage across an inductor at a specific time t. It is dependent on the current and inductance of the inductor.

5. Can you provide an example of solving a circuit with inductors?

Sure, let's say we have a circuit with a 10mH inductor, a 5V voltage source, and a 2Ω resistor in series. Using Kirchhoff's laws, we can set up the equation: 5V - 2Ωi(t) - 10mH(di/dt) = 0. We can then solve for the current i(t) using methods such as integrating the equation or using Laplace transforms.

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