Whiteboard Key for Series/Parallel Circuits
  1. Any reduced circuit drawing. [1 pt]
  2. Any reduced circuit drawing. [1 pt]
  3. Total resistance is 56.7W or 57 W . [1 pt]


    4. R2 and R3 are parallel and must be added first. Let R(23) equal R2 and R3 added together

    1/R(23) = 1/R2 + 1/R3

    1/R(23) = 1/20 ? + 1/10 ?

    1/R(23) = 3/20 ?

    1/R(23) = .15 1/?

    R(23) = 6.7 ?

    Replacing R(23) for R2 and R3 we can now add all resistors in series.

    R1 + R(23) + R4 + R5 = Total Resistance

    20 ? + 6.7 ? + 20 ? + 10 ? = 56.7 ?

    Total Resistance = 56.7 ? or 57 ?

  4. Total current through the battery is 0.18 A. [1 pt]


    5. V = IR

    10 V = I (56.7 ?)

    I = 10 V/56.7 ?

    I = .176 A

  5. R1, R4, R23, and R5, all have .18 A of current.


    6. For R1, R4, and R5

    I(R1) = I(T) = .176 A

    R2 and R3 must be found individually using the Voltage drop across R23:

    V23 = I23*R23

    V23 = .176 A (6.7 ?)

    V23 = 1.18 V

    This voltage drop is the same for R2 and R3, therefore:

    V23 = I2 * R2

    1.18 V = I2 * 20?

    I2 = .059 A

    V23 = I3 * R3

    1.18V = I3 * 10?

    I3 = .118 A
     
     

  6. Voltage drop for each resistor must equal total voltage drop for entire system, which is 10 V.
Therefore:

For R1, and R4 V = IR

V = .176 A * 20?

V = 3.52 V

For R5, V = IR

V = .176 A * 10?

V = 1.76 V

For R23, V = IR

V = .176 A * 6.7?

V = 1.18 V

Because R2 and R3 are in parallel you must use their combined resistance to find the voltage drop over each individual resistor.

For R2, V = IR

V = .059 A * 6.7?

V = .40 V

For R3, V = IR

V = .118 A * 6.7?

V = .80

All voltages add up to the total voltage drop across the battery.

3.52 V + 3.52 V + 1.76V + .40 V + .80 V = 10 V
 

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WHITEBOARD KEY (for original 17Feb00 S/P circuit):

Series/Parallel resistor combination from PHY112 98 MTEx2