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EET 110 Survey of Electronics

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

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  1. EET 110Survey of Electronics Chapter 3 Problems Working with Series Circuits

  2. 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

  3. 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

  4. 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

  5. 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

  6. For figure 3-60 • Voltage across R2 – from above V2 = 4V • Resistance of R3 = V3/I3 R3 = 8/2A = 4 Ω

  7. 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

  8. For figure 3-61 • R3 = V3/I3 = 2.5V/.5A = 5 Ω 15. I3 = .5A

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