1. EET 110 Survey 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 I T = Vt/Rt = 10v/10ohm = 1 amp I 1 = I 2 = I T = 1 amp

4. For Figure 3-59 5 – Current through R2 From above I 2 = I T = 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 = 8 + V2 + 8 or V2 = 20 – 16 = 4v R2 = V2/I2 = 4v/2A = 2 Ω

6. For figure 3-60 9.Voltage across R2 – from above V2 = 4V 10. Resistance of R3 = V3/I3 R3 = 8/2A = 4 Ω

7. For figure 3-61 11. V1 = I1 x R1 =.5A x 1 Ω =.5V =.5V 12. R2 = V2/I2 = 2V/.5A = 4V 13. V3 = VT –(V1 + V2) = 5V – (.5V + 2 V) = 5V – (.5V + 2 V) = 2.5 V = 2.5 V

8. For figure 3-61 14.R3 = V3/I3 = 2.5V/.5A = 5 Ω = 5 Ω 15. I3 =.5A