Solving an Electric Circuit with Kirchhoff's Voltage Law

In summary, an electric circuit with a resistor, capacitor, inductor, and voltage source is described. Using Kirchho 's voltage law, the charge on the capacitor at time t seconds can be represented by the equation d^2q/dt^2 + 6dq/dt + 5q = 26 cos t. The relation between charge and current is explained, with di/dt being the second derivative of charge with respect to time. The problem is solved by plugging in the given values.
  • #1
pokgai
3
0

Homework Statement


An electric circuit consists of a resistor with a resistance of 12 ohms, a capacitor with
a capacitance of 0:1 farads and an inductor with an inductance of 2 henry connected in
series with a voltage source of 52 cos t volts. Initially the charge on the capacitor is 3
coulombs and the current in the circuit is zero.
(a) Using Kirchho 's voltage law, show that the charge q(t) coulombs on the capacitor
at time t seconds satisfi es
d2q/dt2 +6dq/dt+5q = 26 cos t


Homework Equations



V-iR-q/c-Ldi/dt=0

The Attempt at a Solution


It's pretty basic in that you just plug in the values.
52cos(t)-dq/dt12-q/0.1-2di/dt=0
The only thing I don't understand is the relation with di/dt. I am doing calculus and as such I am not familiar with physics. Just wondering how does that di/dt become d^2q/dt^2.

Cheers
 
Physics news on Phys.org
  • #2
The current i is the derivative of the charge q with respect to time:

i=dq/dt.

di/dt is the derivative of i with respect to time, that is

di/dt=d(di/dt)/dt,

it is called "the second derivative" of q and denoted as

d^2d/dt^2.

ehild
 

1. What is Kirchhoff's Voltage Law (KVL)?

Kirchhoff's Voltage Law is a fundamental principle in electrical engineering that states that the sum of all voltages in a closed loop in an electric circuit is equal to zero. This means that the total voltage gained by a charge as it moves around a loop must be equal to the total voltage dropped.

2. How do I apply KVL to solve an electric circuit?

To solve an electric circuit using KVL, you first need to identify all the loops in the circuit. Then, you can write an equation for each loop by summing up all the voltage sources and voltage drops in that loop. Finally, you can solve the resulting system of equations to find the unknown voltages in the circuit.

3. Can KVL be applied to any type of electric circuit?

Yes, KVL can be applied to any type of electric circuit, regardless of its complexity or the types of components it contains. KVL is a general principle that applies to all electric circuits and is a fundamental tool for analyzing and solving them.

4. What are the limitations of KVL?

KVL assumes that there are no magnetic fields present in the circuit and that all components are ideal, meaning they have no resistance or capacitance. In real-world circuits, there may be some resistance and other factors that can affect the accuracy of KVL. Additionally, KVL can only be applied to lumped circuits, where all the components are connected in a single loop.

5. Can KVL be used to calculate the current in a circuit?

No, KVL is used to calculate the voltages in a circuit, not the current. To calculate the current in a circuit, you would need to use Kirchhoff's Current Law (KCL), which states that the sum of all currents entering and leaving a node in a circuit is equal to zero.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
295
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
806
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
26
Views
580
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
202
  • Introductory Physics Homework Help
Replies
5
Views
2K
Back
Top