I need fast R,Q3. An electrical system is shown inthe figure. Where, V is the input, and q,q1 and q3 are the general charges.R.92R3L,a) Write missing terms in energy andvirtual work expressions given below.b) Find the equations of motion(dynamic equation) for q3 and arrangethe equation.E, =..... .......... ..... ................ ..... . ..... .......... ..... . ......... ......... ... .............1 1112.2.E, =2+2 С2 CSW = V 8q -R,4,8q, –R, (4-4,)8(q-q,)...For q3:

Answered: R, Q3. An electrical system is shown in…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

See similar textbooks

Related questions

Question

I need fast

R,
Q3. An electrical system is shown in
the figure. Where, V is the input, and q,
q1 and q3 are the general charges.
R.
92
R3
L,
a) Write missing terms in energy and
virtual work expressions given below.
b) Find the equations of motion
(dynamic equation) for q3 and arrange
the equation.
E, =
..... ....
...... ..... ................ ..... . ..... ....
...... ..... . ......... ...
...... ... ...
..........
1 1
1
1
2.
2.
E, =
2
+
2 С
2 C
SW = V 8q -R,4,8q, –R, (4-4,)8(q-q,)...
For q3:

Expert Solution

Step 1: Solution (a)

In the given circuit, the energy storage elements are 2 capacitors $(C_{1}, C_{2})$ and 2 inductors $(L_{1}, L_{2})$ .

The energy stored in a capacitor can be written as

$E = \frac{1}{2} \frac{1}{C} q^{2}$

and that for an inductor is

$E = \frac{1}{2} L {\dot{q}}^{2} .$

Step 2: Solution (a)

From the circuit, the currents flowing through the inductors $(L_{1}, L_{2})$ are ${\dot{q}}_{2}$ and ${\dot{q}}_{4}$ respectively. The energy stored in these inductors is

$E_{1} = \frac{1}{2} L_{1} {\dot{q}}_{2}^{2} + \frac{1}{2} L_{2} {\dot{q}}_{4}^{2} .$

Answer: $E_{1} = \frac{1}{2} L_{1} {\dot{q}}_{2}^{2} + \frac{1}{2} L_{2} {\dot{q}}_{4}^{2}$ .

The average work is done by only the resistor in an electric circuit, inductors and capacitors do not contribute to it.

Using the concept of Energy and power relation

$P = \frac{δ W}{δ t} δ W = P δ t .$

For an active element, the corresponding work equation is

$δ W = (V \frac{δ q}{δ t}) δ t = V δ q .$

Similarly, the corresponding work equation for a resistor is

$δ W = (R i \cdot i) δ t = (R \dot{q}) \frac{δ q}{δ t} δ t = (R \dot{q}) δ q .$

Step by step

Solved in 4 steps

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.

Similar questions