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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 110 survey of electronics

EET 110Survey of Electronics

Chapter 3 Problems

Working with Series Circuits



For figure 3 59
For figure 3-59

  • 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


For figure 3 591
For Figure 3-59

  • 5 – Current through R2

    From above I2 = IT = 1 amp

  • 6 – voltage across R2

    From Ohms Law V = IR = 1A x 6Ω = 6v


For figure 3 60
For figure 3-60

  • 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


For figure 3 601
For figure 3-60

  • Voltage across R2 – from above V2 = 4V

  • Resistance of R3 = V3/I3

    R3 = 8/2A = 4 Ω


For figure 3 61
For figure 3-61

  • 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


For figure 3 611
For figure 3-61

  • R3 = V3/I3 = 2.5V/.5A

    = 5 Ω

    15. I3 = .5A


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