Combining Resistors In Series In Parallel Example problems Voltage Division Examples Current Division Examples Home Work Problems

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Session 7: Focus
 Combining Resistors
◦ In Series
◦ In Parallel
◦ Example problems
 Voltage Division
◦ Examples
 Current Division
◦ Examples
 Home Work Problems

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Resistors in Series

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Resistors in Series and Parallel
 It is often possible to replace relatively complicated
resistor combinations with a single equivalent resistor
 This is useful when we are not specifically interested in
the current, voltage, or power associated with any of the
individual resistors in the combinations
 All the current, voltage, and power relationships in the
remainder of the circuit will be unchanged

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Resistors in Series
 The current i flowing through all the resistors in series is the same
 Applying KVL we get:
◦ vs = v1 + v2 + v3 …. + vN
 And with Ohm’s law:
◦ vs = iR1 + iR2 + iR3 …. + iRN = i(R1 + R2 + R3 …. + RN)
◦ vs = iRequ
 Requ = R1 + R2 + R3 …. + RN

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S7_Problem_1: Resistors in Series
 Use source and resistor combinations to find i:
Note: This is Example problem 3.11 on page 56, Figure 3.27 in the Ref 1 book (by Hayt)
3 A
90 = i 30
i = 90/30 = 3 A

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S7_HW_Problem_1: Resistors in Series
 Use source and resistor combinations to find i:
Note: This is Practice problem 3.12 on page 57, Figure 3.28 in the Ref 1 book (by Hayt)
-333 mA

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Resistors in Parallel

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Resistors in Parallel
 The voltage v across all the resistors in parallel is the same
 Applying KCL we get:
◦ is = i1 + i2 + i3 …. + iN
 And with Ohm’s law:
◦ is = v/R1 + v/R2 + v/R3 …. + v/RN = v/(R1 + R2 + R3 …. + RN)
◦ vs = v/Requ
 Requ = 1/(R1 + R2 + R3 …. + RN) =

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Resistors in Parallel
 The Requ can be written as well:
 In terms of Conductance, as:
 A parallel combination is routinely indicated by the following
shorthand notation:

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Resistors in Parallel … contd.
 The special case of only two parallel resistors is encountered
fairly often, and is given by
 Or, more simply,
 The last form is worth memorizing, although it is a common error
to attempt to generalize it to more than two resistors

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S7_Problem_2: Resistors in Parallel
 Find v :
Note: This is Practice problem 3.13 on page 58, Figure 3.30 in the Ref 1 book (by Hayt)
50 V
Requ = 10 * 10/(10+10) = 100/20
v = 10 * 5 = 50V
Combine the current source:
i = 5 – 1 + 6 = 10 A
Combine the resistors:
Requ = 5 Ω10 A 5 Ω
i +
-
v

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S7_HW_Problem_2: Resistors in Parallel
 Find vx:
Note: This is Practice problem 3.14 on page 60, Figure 3.32 in the Ref 1 book (by Hayt)
2.819 V

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Resistors in Series – Voltage Division

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Resistors in Series: Voltage Division
 Applying KVL we get:
◦ v(t) = vR1 + vR2 = i(t) R1 + i(t) R2 = i(t) (R1 + R2)
◦ i(t) = v(t) /(R1 + R2)
 With Ohm’s law:
◦ vR1 = i(t)R1 = v(t) R1 / (R1 + R2) = v(t) [R1 / (R1 + R2)]
◦ Similarly
◦ vR2 = i(t)R2 = v(t) R2 / (R1 + R2) = v(t) [R2 / (R1 + R2)]
 This shows the operation of voltage divider
i(t)
v(t)

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Example 1: Voltage Division
 Find vx :
Note: This is Practice problem 3.15 on page 63, Figure 3.36 in the Ref 1 book (by Hayt)
2 V
RPequ = 10 * 10/(10+10) = 100/20
vx = [2/(2+3+5)] * 10 = 2 VCombine parallel resistors:
RPequ = 5 Ω
5 RPequ

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Session 7: Summary
 Combining Resistors
◦ In Series
◦ In Parallel
◦ Example problems
 Voltage Division
◦ Examples
 Home Work Problems

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References
Ref 1 Ref 2