Resistance

1. Unit 2Unit 2
Resistance in Series
and Parallel Circuits
& Complex Circuits
Lesson Modified forLesson Modified for
Grade 9 ScienceGrade 9 Science

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. What is a complex circuit?
Ans. A circuit composed of both series andA circuit composed of both series and
parallel circuits. These circuits can be veryparallel circuits. These circuits can be very
difficult to solve (finddifficult to solve (find Req, all, all V andand I values).values).
Example: Determine: Req, IDetermine: Req, ISS, I, I11, I, I2,2, II3,3, VV1,1, VV2,2, VV33
V S
IS
R 1
R 3
R 2
R 1
= 1 5 .0
R 2 = 5 .0
R 3
= 2 .0
V S
= 1 2 .0 V

7. The hardest part is to find where to start theThe hardest part is to find where to start the
circuit analysis. For the given circuit, thiscircuit analysis. For the given circuit, this
is theis the dotted portion, which is a
________ circuit with respect to the circuit.
The circuit analysis will be done on the
chalkboard.
Answers:
Req = 15/4 Ω +2.0 Ω = 5.75 Ω,
I1 = 0.5 A, I2 =1.6 A, I3 = IS = 2.1 A,
V1 =V2 = 7.8 V, V3 = 4.2 V

8. Practice Questions
1.1. Answer questions on worksheet handout.Answer questions on worksheet handout.
2.2. 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