Series & Parallel Resistors 3/17/2015 1


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1 Series & Parallel Resistors 3/17/2015 1
2 Series Resistors & Voltage Division Consider the singleloop circuit as shown in figure. The two resistors are in series, since the same current i flows in both of them. Applying Ohm s law to each of the resistors, we obtain v 1 = ir 1 and, v 2 = ir 2. (1) If we apply KVL to the loop, we have v + v 1 + v 2 = 0... (2) Combining Eqs. (1) and (2), we get Fig.1 v = v 1 + v 2 = i(r 1 + R 2 )...(3) or i = v / (R 1 + R 2 )..... (4) Notice that Eq. (3) can be written as v = ir eq.....(5) Fig.2 implying that the two resistors can be replaced by an equivalent resistor R eq ; that is, R eq = R 1 + R (6) Thus, Fig.1 can be replaced by the equivalent circuit in Fig. 2. The two circuits in Figs. 1 and 2 are equivalent because they exhibit the same voltagecurrent relationships at the terminals ab. An equivalent circuit such as the one in Fig. 2 is useful in simplifying the analysis of a circuit. 3/17/2015 2
3 In general, For N resistors in series then, To determine the voltage across each resistor in Fig. 1, we substitute Eq. (4) into Eq. (1) and obtain This is called the principle of voltage division, and the circuit in Fig. 1 is called a voltage divider. In general, if a voltage divider has N resistors (R 1,R 2,..., R N ) in series with the source voltage v, the nth resistor (R n ) will have a voltage drop of 3/17/2015 3
4 Parallel Resistors & Current Division Consider the circuit in Fig.1, where two resistors are connected in parallel and therefore have the same voltage across them. From Ohm s law, v = i 1 R 1 = i 2 R 2 or (1) Applying KCL at node a gives the total current i as Substituting Eq. (1) into Eq. (2), we get (2) (3) Fig.1 where Req is the equivalent resistance of the resistors in parallel: (4) 3/17/ (5)
5 We can extend the result in Eq. (4) to the general case of a circuit with N resistors in parallel. The equivalent resistance is (6) It is often more convenient to use conductance rather than resistance when dealing with resistors in parallel. From Eq. (2.38), the equivalent conductance for N resistors in parallel is (7) 3/17/2015 5
6 CURRENT DIVISION i 2 / i 1 = R 1 / R 2 Or, i 2 = i 1 R 1 / R (4) Insert the value of i 1 from eq. (1) Therefore, i 2 = (i i 2 ) R 1 / R 2 i 2 R 2 = i R 1 i 2 R 1 i R 1 = i 2 (R 1 + R 2 ) i = i 1 + i 2 Or, i 1 = i i (1) But, i 2 = v / R (2) and i 1 = v / R (3) From equations (2) and (3), We have: Or, i 2 = i R 1 / (R 1 + R 2 ) (5) Insert the value of i 2 from eq. (5) into eq. (4) we have, i 1 = i R 2 / (R 1 + R 2 ) 3/17/2015 6
7 Example: Find R eq for the circuit shown in Fig. 1. The 6Ω and 3 Ω resistors in the given Fig. 1 are in parallel therefore, R eq = 6 3 / = 2 Ω Also, the 1 Ω and 5 Ω resistors are in series; hence R eq = = 6 Ω Thus the given circuit of Fig. 1 is reduced to that in Fig. 2. In Fig. 2. the two 2 Ω resistors are in series, so R eq = = 4 Ω This 4 Ω resistor is now in parallel with the 6 Ω resistor in Fig. 2; Therefore, R eq = 4 6 / = 2.4 Ω The circuit in Fig. 2 is now replaced with that in Fig. 3. In Fig. 3, the three resistors are in series. Hence, the equivalent resistance for the circuit is Req = = 14.4 Ω Fig. 1 Fig. 2 Fig. 3 3/17/2015 7
8 Example: Find R ab of the given circuit /17/2015 8
9 Example: Find R 13 for the circuit shown in Example: Find R 12 for the circuit shown in 3/17/2015 9
10 Example: Calculate G eq in the circuit of the given Fig. S Therefore, G eq = 12 x 6 / = 4s 3/17/
11 Example: Calculate the equivalent resistance R ab in the circuit in the givenfig. a a b Therefore, R ab = = 11.2 Ohms b 3/17/
12 Example: Find v 1 and v 2 in the circuit shown in the given Fig. Also calculate i 1 and i 2 and the power dissipated in the 12 and 40 Ohms resistors. Total current, i = 15/12 = 1.25 A i 1 = 1.25 x 6/12+6 = A I 2 = 1.25 x 10/10+40 =0.25A V 1 = i 1 x 12 = x 12 = V V 2 = i 2 x 40 = 0.25 x 40 = 10V P 12 = i 1 2 x 12 = (0.416) 2 x 12 = 2W P 40 = i 2 2 x 40 = (0.25) 2 x 40 = 248W 3/17/
13 Example: Find v 1, v 2 and v 3 in the following circuit. 8 3/17/
14 Example: Calculate v 1, i 1, v 2 and i 2 in the given circuit. 3/17/
15 Example: In the given circuit find v, I, and the power absorbed by the 4 Ohm resistor. 3/17/
16 Example: Calculate V O and I O in the following circuit. 3/17/
17 Example: Calculate the equivalent resistance R ab at terminals ab for each of the circuits given below: (a) (b) X 3/17/
18 (C) (D) 3/17/
19 Example: If R eq = 50 Ohms in the given circuit find R. 3/17/
20 Example: Reduce the following circuit shown in the given figure to a single resistor at terminals ab. 3/17/
21 Example: Reduce the following circuit shown in the given figure to a single resistor at terminals ab. 3/17/
22 Example: Calculate the equivalent resistance R ab at terminals ab for the given circuit. 3/17/
23 Example: Find the equivalent resistance R ab in the following circuit: 3/17/
24 TUTORIAL # 01 Q.1 Find v 1 and v 2 in the following circuit. 3/17/
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