Chapter 3: Simple Resistive Circuits
Resistors in Series
When just two elements connect at a single node, they are said to be in series. Series-connected
circuit elements carry the same current. The resistors in the circuit shown in the following
figure are connected in series. We can show that these resistors carry the same current by
applying Kirchhoffs current law to each node in the circuit.
To find is, we apply Kirchhoffs voltage law around the single closed loop. Defining the voltage
across each resistor as a drop in the direction of is gives
the seven resistors can be replaced by a single resistor whose numerical value is the sum of the
individual resistors, that is,
In general, if k resistors are connected in series, the equivalent single resistor has a resistance
equal to the sum of the k resistances, or
Note that the resistance of the equivalent resistor is always larger than that of the largest resistor
in the series connection.
Resistors in Parallel
When two elements connect at a single node pair, they are said to be in parallel. Parallelconnected circuit elements have the same voltage across their terminals. The circuit shown in
the following figure illustrates resistors connected in parallel.
Resistors in parallel can be reduced to a single equivalent resistor using Kirchhoffs current law
and Ohm's law, as we now demonstrate. In the circuit shown, we let the currents i1, i2, i3 an i4
be the currents in the resistors R1 through R4, respectively. We also let the positive reference
direction for each resistor current be down through the resistor, that is, from node a to node b.
From Kirchhoffs current law,
The parallel connection of the resistors means that the voltage across each resistor must be the
same. Hence, from Ohm's law,
We can see that the four resistors in the circuit shown can be replaced by a single equivalent
resistor. The circuit shown above illustrates the substitution.
For k resistors connected in parallel
Note that the resistance of the equivalent resistor is always smaller than the resistance of the
smallest resistor in the parallel connection.
Many times only two resistors are connected in parallel:
We calculate the equivalent resistance
Don't make the mistake of assuming that two elements are parallel connected merely because
they are lined up in parallel in a circuit diagram. The defining characteristic of parallelconnected elements is that they have the same voltage across their terminals. In the following
figure, you can see that R1 and R3 are not parallel connected because, between their respective
terminals, another resistor dissipates some of the voltage.
The Voltage-Divider and Current-Divider Circuits
Voltage-Divider
At times—especially in electronic circuits—developing more than one voltage level from a
single voltage supply is necessary. One way of doing this is by using a voltage-divider circuit,
such as the one in the following figure:
From Kirchhoffs current law KCL, R1 and R2 carry the same current. Applying Kirchhoffs
voltage law around the closed loop yields
vs = iR1 + iR2,
Now we can use Ohm's law to calculate v1 and v2:
This show that v1 and v2 are fractions of vs. Each fraction is the ratio of the resistance across
which the divided voltage is defined to the sum of the two resistances. Because this ratio is
always less than 1, the divided voltages v1 and v2 are always less than the source voltage vs.
The generalization will yield a very useful circuit analysis technique known as voltage
division.
For voltage division:
If we have n resistors connected in series.
We are interested in finding the voltage drop vj across an arbitrary resistor Rj in terms of the
voltage v
The Current-Divider Circuit
The current-divider circuit shown in Figure below consists of two resistors connected in
parallel across a current source. The current divider is designed to divide the current is between
R1 and R2.
We find the relationship between the current is and the current in each resistor (that is, i1 and
i2) by directly applying Ohm's law and Kirchhoffs current law. The voltage across the parallel
resistors is:
The generalization will yield a very useful circuit analysis technique current division.
For current division:
We are interested in finding the current ij through an arbitrary resistor Rj in terms of the current
i. We start by using Ohm's law to calculate v, the voltage drop across each of the resistors in
parallel, in terms of the current i and the n resistors:
In summary we can write for the generalization of the current divider:
Measuring Resistance— The Wheatstone Bridge
Many different circuit configurations are used to measure resistance. Here we will focus on
just one, the Wheatstone bridge. The Wheatstone bridge circuit is used to precisely measure
resistances of medium values, that is, in the range of 1 Ω to 1 MΩ. The bridge circuit consists
of four resistors, a dc voltage source, and a detector. The resistance of one of the four resistors
can be varied.
R1 R2, and R3 are known resistors and Rx is the unknown resistor.
To find the value of Rx, we adjust the variable resistor R3 until there is no current in the
galvanometer. We then calculate the unknown resistor from the simple expression
The derivation of this equation follows directly from the application of Kirchhoff s laws to the
bridge circuit. When ig is zero, that is, when the bridge is balanced, Kirchhoffs current law
requires that
Now, because ig is zero, there is no voltage drop across the detector, and therefore points a and
b are at the same potential. Thus when the bridge is balanced, Kirchhoff s voltage law requires
that
=>
Delta-to-Wye (Pi-to-Tee) Equivalent Circuits
Exercises
Ex 1
For the circuit shown, find (a) the voltage v,
(b) the power delivered to the circuit by the current source,
(c) the power dissipated in the 10 ohms resistor.
Solution
Start from the right hand side of the circuit and make series and parallel combinations of the
resistors until one equivalent resistor remains. Begin by combining the 6 Ω resistor and the 10
Ω resistor in series:
6 Ω + 10 Ω = 16 Ω
Now combine this 16 Ω resistor in parallel with the 64 Ω resistor:
This equivalent 12.8 Ω resistor is in series with the 7.2 Ω resistor:
12.8 Ω + 7.2 Ω = 20 Ω
Finally, this equivalent 20 Ω resistor is in parallel with the 30 Ω resistor:
a] With the simplified circuit we can use Ohm’s law to find the voltage across both the
current source and the 12 Ω equivalent resistor:
v = (12 Ω)(5 A) = 60 V
[b] Now that we know the value of the voltage drop across the current source, we can use the
formula p = −vi to find the power associated with the source:
p = −(60 V)(5 A) = −300 W
Thus, the source delivers 300 W of power to the circuit.
c] We now can return to the original circuit, shown in the first figure. In this circuit, v = 60 V,
as calculated in part (a). This is also the voltage drop across the 30 Ω resistor, so we can use
Ohm’s law to calculate the current through this resistor:
Now write a KCL equation at the upper left node to find the current iB:
Next, write a KVL equation around the outer loop of the circuit, using Ohm’s law to express
the voltage drop across the resistors in terms of the current through the resistors:
Now that we have the current through the 10 Ω resistor we can use the formula p = Ri2 to find
the power:
Other methods to calculate ic
-KVL middle path loop: 60 volts= V64 + vB,
OHM: vB= RB.iB = 7.2x3= 21.6 V
→ V64= 60 – 21.6 = 38.4 Volts ; V64 = V16 → ic = 38.4 / 16 = 2.4 A
-Or we can calculate i64 : V64 = 64 . i64 → i64 = 38.4 / 64 = 0.6 A
KCL: ic = iB - i64 = 3 – 0.6 = 2.4 A
-Or we can use current divider method example: ic =
//
× 𝑖𝐵 =
.
× 3 = 2.4 𝐴
Ex2 Using Voltage Division and Current Division to Solve a Circuit
Use current division to find the current i0 and use voltage division to find the voltage v0 for
the circuit
This is also the voltage drop across the branch containing the 40 Ω, the 10 Ω, and the 30 Ω
resistors in series. We can then use voltage division to determine the voltage drop v0 across
the 30 Ω resistor given that we know the voltage drop across the series connected resistors, To
do this, we recognize that the equivalent resistance of the series-connected resistors is 40 + 10
+ 30 = 80 Ω
Ex3
In the following figure, resistor RL acts as a load on the circuit. A load on any circuit consists
of one or more circuit elements that absorb energy from the circuit. We have vs = 20 V and
R1 = 40 ohms.
a. Given v0 = 4V when no load is applied, find R2.
b. Find RL when v0 = 3V.
Solution:
a) v0 = vs . R2 / (40 + R2) =20 ×
= 4→R2=10 ohms.
b) Re=R2 // RL = R2.RL/ (R2+RL) ; R2=10 ohms => Re=
v0 = vs . Re/(40+Re) =20 ×
=> Re =
= 3 => 20 Re = 3 Re +120 => 17 Re=120
=
→RL = 24 Ω.
Ex 4
a. Use the voltage division to find v0.
b. Determine the current i40 in the 40 Ω resistor and the current i30 in the 30 Ω resistor.
c. What is the power absorbed by the 50 Ω resistor?
Solution
a)
Now we can use voltage division to find the voltage vo:
b)
or
Current division to calculate i30 :
c) Current division to calculate i50 :
Ex 5
In the circuit,
a.
b.
c.
d.
Find v
Find i0
Find 𝑃12Ω
Find P12A
a)
Current divider : i1=
. 12 or we can write also : i1=
/
/
. 12
R1=40+10+20=70 Ω
Req= 70 // (12+ (20//180)) = 70 // (12+
i1=
×
. 12 = i1= . 12 =3.6 A
you can also write :
To find v : v = 40. i1 = 40 × 3.6 = 144 𝑉
) =70 // (12+ 18) = 70//30 =
×
=21
b)
We can also write : 180//20=18 and i0 =
×𝑖 =
× 8.4 = 7.56𝐴
c) P12 Ω = 12 × 8.4 = 846.72𝑊
d) P12A = - 12× 𝑣
= −12 × 𝑅 × 𝑖 = -12× 70 × 3.6 = −3024𝑊
Ex 6: Find v1 and v2
Solution
Req=(50+25)// [30+(30//60)]=75//[30+
Req=75//50=
×
×
]=75//(30+20)=>
=30
current divider: i2 = i1 .
//
we can write it also as i2 = i1×
= i1 .
//
= 25 ×
= 25×
= 15A.
//
=15A.
//
v2= 15A× ( 30//60)= 15 × 20 =300 V
v1= 25× (𝑅𝑒𝑞 + 12)= 25× (30 + 12)= 25 × 42= 1050 V.
Ex 7 find the current and the power delivered by the 40 V source.
Ex 8
Use 𝒀 − ∆ to find v
Solution
Convert the three Y-connected resistors, 20Ω, 10Ω, and 5Ω to three ∆-connected resistors
Ra,Rb, and Rc. To assist you the figure below has both the Y-connected resistors and the ∆connected resistors
From this circuit we see that the 70Ω resistor is parallel to the 28Ω resistor:
Also, the 17.5Ω resistor is parallel to the 105Ω resistor:
Once the parallel combinations are made, we can see that the equivalent 20Ω resistor is in
series with the equivalent 15Ω resistor, giving an equivalent resistance of
Finally, this equivalent 35Ω resistor is in parallel with the other 35Ω resistor:
In summary: