Assignment (C288)

solving circuit information using Ohm's Law
V=IR
1. Series Circuit: V=12V, R1=3 Ohms, R2=9 Ohms
        a. circuit current I=V/(R1+R2)=12 V / (3 Ohms +9 Ohms) = 12/12= 1 A
        b. Voltage drop across R1. V=IR1=1A x 3 Ohms = 3V
        c. Voltage drop across R2. V=IR2= 1A x 9 Ohms = 9V
        d. Total Resistance. Rt=R1+R2 = 3 Ohms + 9 Ohms = 12 Ohms
        e. Diagram

2. Parallel Circuit
         V=12 V, Branch 1: R1=2 Ohms and Branch 2: R2 4 Ohms
            a. Circuit Current: I=V/R; 1/Rt=1/R1+1/R2=3/4, Rt=4/3, I=12x3/4=9 A
            b. Voltage drop across R1 = 12 V
            c. Voltage drop across R2 = 12 V
            d. Circuit Resistance: Rt=R1R2/(R1+R2)=2x4/(2+4)=1.33 Ohms
            e. Current through R1: I1=V/R1=12V/2 Ohms=6 A
            f. Current through R2: I2=V/R2= 12/4=3 A
           g. Circuit Diagram

3. Series-Parallel (Compound) Circuits

        V=12 V, R1=2 Ohms in series with parallel of R2=3 Ohms and R3 = 3 Ohms

            a. Circuit Current: I=V/Rt; Rt=R2R3/(R2+R3)+R1=1.5+2=3.5 Ohms; I=12/3.5=3.43 A

            b. Voltage Drop across R1: 3.43 A x 2= 6.86V

            c. Voltage Drop across R2 and R3: 3.43A x 1.5 Ohms = 4.145 V

            d. Total Circuit resistance: Rt= R2R3/(R2+R3)+R1=9/6+2=3.5 Ohms

            e. Circuit diagram: