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# EET 110 Survey of Electronics - PowerPoint PPT Presentation

EET 110 Survey of Electronics. Chapter 3 Problems Working with Series Circuits. 1 – Current for Figure 3-58 From OHMS LAW I=V/R I = 10v/5 Ω = 2 Amps 2 – Voltage across R (VR) = 10 V. For figure 3-59. 3- Total resistance R Rt = R1 + R2 = 4Ω + 6 Ω = 10 ohm 4- Current through R1

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### EET 110Survey of Electronics

Chapter 3 Problems

Working with Series Circuits

• 3- Total resistance R

Rt = R1 + R2 = 4Ω + 6 Ω = 10 ohm

• 4- Current through R1

IT = Vt/Rt = 10v/10ohm = 1 amp

I1 = I2 = IT= 1 amp

• 5 – Current through R2

From above I2 = IT = 1 amp

• 6 – voltage across R2

From Ohms Law V = IR = 1A x 6Ω = 6v

• 7. V for R1

Given IT = 2 A

V(R1) = IxR = 2A x 4 Ω = 8 V

• 8. Resistance of R2

If we know V(R1) = 8 V and V(R3) = 8V

VT=20= V1 + V2 + V3

= 8 + V2 + 8 or

V2 = 20 – 16 = 4v

R2 = V2/I2 = 4v/2A = 2 Ω

• Voltage across R2 – from above V2 = 4V

• Resistance of R3 = V3/I3

R3 = 8/2A = 4 Ω

• V1 = I1 x R1 = .5A x 1 Ω

= .5V

• R2 = V2/I2 = 2V/.5A = 4V

• V3 = VT –(V1 + V2)

= 5V – (.5V + 2 V)

= 2.5 V

• R3 = V3/I3 = 2.5V/.5A

= 5 Ω

15. I3 = .5A