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Unit 3 –Lesson 4Unit 3 –Lesson 4
Resistance in Series
and Parallel Circuits
Nelson Reference Pages:Nelson Reference Pages:
527 – 530527 – 530
All loads in a circuit willAll loads in a circuit will
have resistance. Anhave resistance. An
equivalent (equivalent (RReqeq), or), or total
((RRTT), resistance, for the), resistance, for the
entire circuit, can beentire circuit, can be
calculated. To do this,calculated. To do this,
we assume that allwe assume that all
resistancesresistances must obey
Ohm’s Law.Ohm’s Law.
For a series circuit:
VS = V1+ V2 +V3
But:
VS =ISReq , V1 =ISR1,
V2 =ISR2 , V3 =ISR3
V 1
V 3
V 2
V S
IS R 1
R 3
R 2
Sub values forSub values for VV into theinto the
first equation to get:first equation to get:
ISReq = ISR1+ ISR2 + ISR3
Simplify to get:
Req = R1+ R2 + R3
To summarize: For “n” resistances in series:To summarize: For “n” resistances in series:
Req = R1+ R2 + …+ Rn , n = 1, 2, 3,…
Resistance in Parallel Circuits
For a parallel circuit, potential difference (voltage)For a parallel circuit, potential difference (voltage)
across each branch is constant.across each branch is constant.
We know that:We know that: IS = I1+ I2 +I3 ,{Kirchhoff’s Node Law}
But: IS =VS /Req , I1 =VS /R1 , I2 =VS /R2 , I3 =VS /R3
Sub in to first equation:
VS /Req =VS /R1 +VS /R2 +VS /R3 and then simplify: 1/Req =1/R1
+1/R2 +1/R3
R 1 R 3
R 2
A 1
V S
I S
A 2 A 3
I1
I2 I3
A S
To summarize: For “n” resistances inTo summarize: For “n” resistances in
parallel:parallel:
Req = (1/R1+ 1/R2 + …+ 1/Rn )-1
Where n = 1, 2, 3,…
Notice the exponent which means get
the reciprocal of the value in the
brackets.
Practice Questions
Nelson Textbook: Page 530 #4, 5
Workbook:
Page 49Page 49 (New 54) # 1, 2, 3, 5, (6a optional)# 1, 2, 3, 5, (6a optional)
Question from McGraw-Hill Physics 11
1.1. A current ofA current of 4.80 A leaves a battery andleaves a battery and
separates into three currents running throughseparates into three currents running through
three parallel loads. The current to the firstthree parallel loads. The current to the first
load isload is 2.50 A, current through the second load, current through the second load
isis 1.80 A, and the resistance of the third load is, and the resistance of the third load is
108 Ohm. Calculate:. Calculate: Req , R1 and R2
Ans. 11.2 Ω, 21.6 Ω, 30.0 Ω
Practice Questions
Nelson Textbook: Page 530 #4, 5
Workbook:
Page 49Page 49 (New 54) # 1, 2, 3, 5, (6a optional)# 1, 2, 3, 5, (6a optional)
Question from McGraw-Hill Physics 11
1.1. A current ofA current of 4.80 A leaves a battery andleaves a battery and
separates into three currents running throughseparates into three currents running through
three parallel loads. The current to the firstthree parallel loads. The current to the first
load isload is 2.50 A, current through the second load, current through the second load
isis 1.80 A, and the resistance of the third load is, and the resistance of the third load is
108 Ohm. Calculate:. Calculate: Req , R1 and R2
Ans. 11.2 Ω, 21.6 Ω, 30.0 Ω

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Grade 11,U3 L4-resistance in Series and Parallel CCT's

  • 1. Unit 3 –Lesson 4Unit 3 –Lesson 4 Resistance in Series and Parallel Circuits Nelson Reference Pages:Nelson Reference Pages: 527 – 530527 – 530
  • 2. All loads in a circuit willAll loads in a circuit will have resistance. Anhave resistance. An equivalent (equivalent (RReqeq), or), or total ((RRTT), resistance, for the), resistance, for the entire circuit, can beentire circuit, can be calculated. To do this,calculated. To do this, we assume that allwe assume that all resistancesresistances must obey Ohm’s Law.Ohm’s Law. For a series circuit: VS = V1+ V2 +V3 But: VS =ISReq , V1 =ISR1, V2 =ISR2 , V3 =ISR3 V 1 V 3 V 2 V S IS R 1 R 3 R 2 Sub values forSub values for VV into theinto the first equation to get:first equation to get: ISReq = ISR1+ ISR2 + ISR3 Simplify to get: Req = R1+ R2 + R3
  • 3. To summarize: For “n” resistances in series:To summarize: For “n” resistances in series: Req = R1+ R2 + …+ Rn , n = 1, 2, 3,…
  • 4. Resistance in Parallel Circuits For a parallel circuit, potential difference (voltage)For a parallel circuit, potential difference (voltage) across each branch is constant.across each branch is constant. We know that:We know that: IS = I1+ I2 +I3 ,{Kirchhoff’s Node Law} But: IS =VS /Req , I1 =VS /R1 , I2 =VS /R2 , I3 =VS /R3 Sub in to first equation: VS /Req =VS /R1 +VS /R2 +VS /R3 and then simplify: 1/Req =1/R1 +1/R2 +1/R3 R 1 R 3 R 2 A 1 V S I S A 2 A 3 I1 I2 I3 A S
  • 5. To summarize: For “n” resistances inTo summarize: For “n” resistances in parallel:parallel: Req = (1/R1+ 1/R2 + …+ 1/Rn )-1 Where n = 1, 2, 3,… Notice the exponent which means get the reciprocal of the value in the brackets.
  • 6. Practice Questions Nelson Textbook: Page 530 #4, 5 Workbook: Page 49Page 49 (New 54) # 1, 2, 3, 5, (6a optional)# 1, 2, 3, 5, (6a optional) Question from McGraw-Hill Physics 11 1.1. A current ofA current of 4.80 A leaves a battery andleaves a battery and separates into three currents running throughseparates into three currents running through three parallel loads. The current to the firstthree parallel loads. The current to the first load isload is 2.50 A, current through the second load, current through the second load isis 1.80 A, and the resistance of the third load is, and the resistance of the third load is 108 Ohm. Calculate:. Calculate: Req , R1 and R2 Ans. 11.2 Ω, 21.6 Ω, 30.0 Ω
  • 7. Practice Questions Nelson Textbook: Page 530 #4, 5 Workbook: Page 49Page 49 (New 54) # 1, 2, 3, 5, (6a optional)# 1, 2, 3, 5, (6a optional) Question from McGraw-Hill Physics 11 1.1. A current ofA current of 4.80 A leaves a battery andleaves a battery and separates into three currents running throughseparates into three currents running through three parallel loads. The current to the firstthree parallel loads. The current to the first load isload is 2.50 A, current through the second load, current through the second load isis 1.80 A, and the resistance of the third load is, and the resistance of the third load is 108 Ohm. Calculate:. Calculate: Req , R1 and R2 Ans. 11.2 Ω, 21.6 Ω, 30.0 Ω