EE 204
Lecture 06
VDR, CDR & Circuit Solution by KVL and KCL
The Voltage Divider Rule (VDR)
The total voltage across the series resistors R1 , R2 , ... , RN is V.
i=
V
V
= N
Req
∑R
i =1
i
⎛
⎜ R
V
vx = iRx = N
Rx = ⎜ N x
⎜ R
Ri
∑
⎜∑ i
i =1
⎝ i =1
VDR ⇒
VDR
⇒
⎛
⎜ R
vx = ⎜ N x
⎜ R
⎜∑ i
⎝ i =1
vresistor =
⎞
⎟
⎟V
⎟
⎟
⎠
⎞
⎟
⎟V
⎟
⎟
⎠
resistor
× (total voltage)
sum
The VDR is valid for any number of resistors in series.
Figure 10
Example 3:
Calculate v1 & v2
VDR ⇒
v1 =
4
× 30 = 12V
4+6
VDR ⇒
v2 =
6
× 30 = 18V
4+6
Ω
Ω
Figure 11
VDR
⇒
Higher voltage drop across the higher resistance.
Example 4:
Calculate the unknown voltages.
Ω
Ω
Ω
Ω
Figure 12
Solution:
5 + 7 = 12Ω
VDR ⇒
VDR
⇒
⇒
R1 =
v1 = −
v2 =
15
× 36 (a minus sign is required here. Why?) ⇒
15 + 3
3
× 36
15 + 3
Check: KVL ⇒
VDR
⇒
4 × 12
= 3Ω
4 + 12
v3 =
⇒
v2 = 6V
−36 − v1 + v2 = −36 − (−30) + (6) = −36 + 30 + 6 = 0
5
5
× v2 = × 6
5+7
12
⇒
v3 = 2.5V
v1 = −30V
⇒
VDR
v4 =
7
7
× v2 = × 6
5+7
12
v4 = 3.5V
⇒
Ω
Ω
Ω
Ω
Ω
Ω
Figure 13
The Current Divider Rule (CDR)
The total current entering into the parallel combination of R1 & R2 is I
V = IReq = I
R1 R2
R1 + R2
V
V
& I2 =
R1
R2
1
RR
R2
I1 = × 1 2 I
⇒ I1 =
I
R1 R1 + R2
R1 + R2
I1 =
⇒ I2 =
Similarly
CDR
⇒
I=
R1
I
R1 + R2
(1)
(2)
other resistor
× total current
sum
CDR applies to only two resistors in parallel.
Figure 14
Example 5:
a) Use CDR to calculate I1 & I 2 .
b) Verify your results by checking KCL.
Ω
Ω
Figure 15
Solution:
a)
I1 =
2
×3
2+5
⇒
I2 =
5
×3
2+5
⇒
I1 =
6
A
7
I2 =
15
A
7
b)
KCL at node “a”
⇒
6 15
21
I s − I1 − I 2 = 3 − − = 3 − = 0
7 7
7
Ω
Ω
Figure 16
CDR
⇒
Higher current passes through the lower resistance.
(KCL verified)
Example 6:
Use CDR to calculate I1 , I 2 & I 3 .
Ω
Ω
Ω
Ω
Figure 17
Solution:
6Ω & 12Ω (in parallel) ⇒
4Ω & 4Ω (in parallel)
⇒
6 × 12 72
=
= 4Ω
6 + 12 18
4 × 4 16
=
= 2Ω
4+4 8
∴ Req = 8 + 2 = 10Ω
I=
20
= 2A
10
4
× 2 = 1A
4+4
CDR
⇒
I4 =
CDR
⇒
I3 = −
CDR
⇒
I1 =
∴ I1 =
CDR
4
× 2 = −1 A
4+4
(the minus sign is necessary in this case. Why?)
12
× I 4 (because I 4 is the total current through 6Ω & 12Ω )
6 + 12
12
2
×1 = A
18
3
⇒
I2 =
6
1
×1 = A
6 + 12
3
Check KCL at super node ⇒
verified)
2 1
I − I1 − I 2 + I 3 = 2 − − + (−1) = 1 − 1 = 0 (KCL
3 3
Verify that the three parallel resistors 6Ω , 12Ω and 4Ω have the same voltage.
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Figure 18
Solutions for Circuits Containing More than One Source
For Circuit with more than one source, we systematically apply KVL, KCL and
Ohm’s Law. It is obvious by now that:
Currents through elements in series are equal. (Referred to as simple
node)
Voltages across elements in parallel are equal. (Referred to as simple loop)
Thus
Apply KCL only for nonsimple nodes, ie for nodes connecting more than
two elements. (Elements not in series).
Apply KVL only for nonsimple loops, ie for loops whose elements are not
connected at both pairs of nodes. (Elements not in parallel)
Example 8
Determine the current i in the circuit of
V1
+
a
i
20
i2
+
V2
20V
0.8 A
+
V2
40
Solution:
−i2 − i + 54 = 0
i2 = 53 A
−v2 + v1 + 20 = 0
−(40 × 53 ) + 20i + 20 = 0
i = 15 A
Example 8
Determine the current i in the circuit
V1
20
+
a
i
+
20V
i2
V2
0.8 A
+
V2
40
Solution:
i = 15 A
i2 = 53 A
Example 9
Determine the voltage V and current I in the circuit
Solution:
We don’t know the voltage or current associated with each resistor, so we cannot
apply Ohm’s law or KVL. Hence we apply KCL at the single nonsimple node a
to obtain the current through the 3-V source in terms of I as I-2 as shown in Fig.
Next apply Ohm’s law to all resistors (in terms of I) then apply KVL around the
outside loop for which we know the voltages across all elements. This yields:
−3V = 2( I − 2) + 2 I + 2 I
Solving yields:
I = 76 A
Apply KVL we obtain V:
V = 2I + 2I
V = 3V − 2( I − 2)
V = 143 V