Unformatted text preview: SERIES
140 IN SERIES de CIRCUITS
which for Fig. 5.12 results in
VI = 1,Ri = 1,R, = (60 mA )(10 0) - 0.6 V
Vy " lykg - 1,ky - (60 mA )(30 0) = 18 V
R, $ 100 0
V, " lyk, = 1,Ry = (60 mA )(100 0) - 6.0 V
Note that in all the numerical calculations appearing in the text thus
far, a unit of measurement has been applied to each calculated quantity.
Always remember that a quantity without a unit of measurement is often
Resistance "seen " at the terminals of a series circuit.
EXAMPLE 5.4 For the series circuit in Fig. 5.15.
resistance "seen" at the connection terminals, as shown in Fig. 5.13(x
In other words, it reduces the entire configuration to one such as in Fig
a Find the total resistance Ry
. Calculate the resulting source current ly
5. 13(b) to which Ohm's law can easily be applied.
For the configuration in Fig. 5.12, with the total resistance calculated
c. Determine the voltage across each resistor
in the last section, the resulting current is
E 8.4 V
= 0.06 A = 60 mA
Series circuit to be investigated in Example 5.4.
Note that the current I, at every point or corner of the network is the
RT - 80
same. Furthermore, note that the current is also indicated on the current
display of the power supply.
b. I m E 20 V
RT 802 25 A
Now that we have the current level, we can calculate the voltage
C. V1 = 1,R, - 1,R1 - (25 A)(2 0) - 5V
across each resistor. First recognize that
Va = 12R2 - 1,Rx - (25 A)(1 0) = 25V
the polarity of the voltage across a resistor is determined by the direction
of the current.
VJ = 13R3 - 1,R, - (25 AX(5 1) - 125V
Current entering a resistor creates a drop in voltage with the polarity
indicated in Fig. 5.14(a). Reverse the direction of the current, and the
EXAMPLE 5.5 For the series circuit in Fig. 5.16:
polarity will reverse as shown in Fig. 5.14(b). Change the orientation of
a. Find the total resistance RT.
the resistor, and the same rules apply as shown in Fig. 5.14(c). Applying
b. Determine the source current /, and indicate its direction on the circuit.
the above to the circuit in Fig. 5.12 will result in the polarities appearing
in that figure.
c. Find the voltage across resistor Ry and indicate its polarity on the
3. The elements of the circuit are rearranged as shown in Fig. 5.17.
Series circuit to be analyzed in Example S.5.
v Z 10 0
RT = R2 + NR
= 40 + (3)(70)
= 40 + 210
( b )
RT = 25 0
Inserting the polarities across a resistor as determined by the direction
of the current.
The magnitude of the voltage drop across each resistor can then be
found by applying Ohm's law using only the resistance of each resistor.
E = 50 V =
VI = IRI
V1 = 12R2
VJ = 13R3
Circuit in Fig. 5.16 redrawn to permit the use of Eq. (5.2)....
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- Fall '19
- SK Abid