Unit 1

Question 1

How is KVL applied here

Answer 1

Clockwise loop across the first set of elements and the second

-V3 + V1 - V4 - V6 = 0

Question 2

Simplify this Circuit

Answer 2

Resistors are in series, we can use:

Re = R1 + R2

Question 3

Write the node analysis for Node (1)

Answer 3

-ia + (V1-V2)/R2 + (V1-V3)/R1 = 0

Question 4

How can we tell if energy is being Supplied for Absorbed by an Element ?

Answer 4

Supplied = negative (-)

Absorbed = positive (+)

Question 5

What is the equation to Power in circuits ?

Answer 5

P = V * i
P = (i^2)R

Question 6

Write the KVL equations

Answer 6

Loop (1): -5V + Va + 10V = 0

Loop (2): Vc - 15V - Va = 0

Loop (3): -10V + 15V + Vb = 0

Question 7

Explain how to solve for power in each element

Answer 7

We can start by using KCL and solve for the current across each element:

Node (1): ic -ia + id = 0
Node (2) -ib -ic - id = 0

Then we can use KVL to solve for the Voltage in each element:

Loop(1): -Vc + Vd = 0
Loop(2): -Va - Vb + Vc = 0

Then we can multiply each Voltage by the Current and get the Power

Question 8

Explain how to solve for iR, Voltage though R and and C

Answer 8

There are no Nodes so iR = 2A

The Voltage of R is V_R = IR
V_R = (2A)(5ohms) = 10V

We can now use KVL to solve for Voltage through C

VC + VR + VV = 0
VC = -20V

Question 9

If the Power through 8omhs Resistor is 8W solve for the Vx

Answer 9

Start by solving that the Current though the left side must be 1A since

P = i²R
8W = i²(8ohms)
i = 1A

So now we can solve for Vin using KVL

-Vin + 2V + 8V = 0

To find Vx we use:

(Vx / 5kohms + 10Kohms) = (Vin / 10ohms)

and solve for Vx

Question 10

How do we use KVL for Resistors ?

Answer 10

V = R₁(i) + R₂(i) + R₃(i)

Re = R₁ + R₂ + R₃

V = Re(i)