Sample Problems - Solutions KVL, KCL & Power

1) For the circuit shown below, determine the voltage for each of the resistors and label the values on the diagram. KVL: ∑Vrises = ∑Vdrops Assume IS clockwise: 10 V = 20 IS + 10 IS + 20 V + 30 IS (note: if all resistors kΩ, drop k and remember answer will be in mA) Solving: -10 V = 60 IS => IS = -1/6 mA (counter clockwise) Applying Ohms Law, V20k = 20/6 = 10/3 V V10k = 10/6 = 5/3 V V20k = 30/6 = 5 V Now, label diagram with voltage drops: 2) In the circuit shown below, the current through the 3 kΩ resistor is measured to be 4 mA. Determine the source voltage (VS) and source current (IS). V3k = V12k = (4 mA)(3kΩ) = 12V I12k = (12 V)/(12 kΩ) = 1 mA IS = I3k + I12k = 5 mA VS = V2k + V3k (KVL) or V2k + V12k = (2 kΩ)(5 mA) + 12 V = 22 V

+-

IS

V S 12 kΩ 3 kΩ

2 kΩ 4 mA

+-10 V +- 20 V

10 kΩ

30 kΩ

20 kΩ

- 1.67 V +

+ 5.0 V -

- 3.33 V +

+-10 V +- 20 V

10 kΩ

30 kΩ

20 kΩ

3) KVL, KCL 3a) For the circuit shown above, determine the voltages VA, VB, VC, and VD indicated on the diagram. Apply KVL to mesh and loop currents, IA, IB, IC, and ID . For loop or Mesh: ∑VRises = ∑VDrops IA: 6 V + 6 V = VA >>> VA = 12 V IB: 28 V + VB = 12 V + 6 V >>> VB = -10 V IC: 12 V + VC = 6 V >>> VC = -6 V ID: 28 V = 12 V + VD >>> VD = 16 V

α

β

X

Y

χ

δ

ε

φI1

I2

I4

I328 V

+

- 4 mA

+ VB -

VA

+

-

6 V+

-

+ 12 V - + 6 V -

VD

+

-

- VC +

1 mA

4 mA

2 mA

IC

ID

IA

IB

3b) Determine the currents I1, I2, I3 and I4 indicated on the diagram. Apply KCL @ node N1, N2, N3, and N4, ∑IIN = ∑IOUT N1: 4 mA = I1 + 1 mA >>> I1 = 3 mA N2: I1 (= 3 mA) = I2 + 4 mA >>> I2 = -1 mA N4: 4 mA = I4 + 2 mA >>> I4 = 2 mA N3: I4 (= 2 mA) + I3 = 4 mA >>> I3 = 2 mA

α

β

X

Y

χ

δ

ε

φI1

I2

I4

I328 V

+

- 4 mA

+ VB -

VA

+

-

6 V+

-

+ 12 V - + 6 V -

VD

+

-

- VC +

1 mA

4 mA

2 mAN1

N2

N3

N4

3c) Determine the power (supplied or dissipated) by elements X & Y. Putting values we solved for in parts a) and b) onto the diagram, we have:

PX = IX VX = (3 mA)(6 V) = 18 mW (Dissipated) Note: Actual current flows from + to - terminals of voltage on the device. PY = IY VY = (2 mA)(10 V) = 20 mW (Dissipated) Note: Actual current flows from + to - terminals of voltage on the device.

X

+ 6 V -3 mAY

- 10 V + 2 mA