Solution:

← Back to Fundamentals of Electric Circuits Overview For e xample, if four 100- Ω resistors are connected in parallel, their equivalent resistance is 25 Ω.

It is often more convenient to use conductance rather than resistance when dealing with resistors in parallel. From Eq. (2.38), the equi valent conductance for N resistors in parallel is

$$G_{\text{eq}} = G_1 + G_2 + G_3 + \dots + G_N \quad (2.40)$$

where Geq = 1/Req, G¹ = 1/R1, G² = 1/R2, G3 = 1/ R3, … , GN = 1/ RN. Equation (2.40) states:

The equivalent conductance of resistors connected in parallel is the sum of their individual conductances.

This means that we may replace the circuit in Fig. 2.31 with that in Fig. 2.32. Notice the similarity between Eqs. (2.30) and (2.40). The equivalent conductance of parallel resistors is obtained the same w ay as the equivalent resistance of series resistors. In the same manner , the equivalent conductance of resistors in series is obtained just the same Conductances in parallel behave as a single conductance whose value is equal to the sum of the individual conductances.

Figure 2.32 Equivalent circuit to Fig. 2.31.

way as the resistance of resistors in parallel. Thus, the equivalent conductance Geq of N resistors in series (such as shown in Fig. 2.29) is

$$\frac{1}{G_{\text{eq}}} = \frac{1}{G_1} + \frac{1}{G_2} + \frac{1}{G_3} + \dots + \frac{1}{G_N}$$

(2.41)

Given the total current i entering node a in Fig. 2.31, ho w do we obtain current i1 and i2? We know that the equi valent resistor has the same voltage, or

$$v = iR_{\text{eq}} = \frac{iR_1R_2}{R_1 + R_2} \tag{2.42}$$

Combining Eqs. (2.33) and (2.42) results in

$$i_1 = \frac{R_2 i}{R_1 + R_2}, \qquad i_2 = \frac{R_1 i}{R_1 + R_2}$$

(2.43)

which shows that the total current i is shared by the resistors in in verse proportion to their resistances. This is known as the principle of current division, and the circuit in Fig. 2.31 is known as a current divider. Notice that the lar ger current flows through the smaller re sistance.

As an extreme case, suppose one of the resistors in Fig. 2.31 is zero, say R² = 0; that is, R2 is a short circuit, as sho wn in Fig. 2.33(a). From Eq. (2.43), R² = 0 implies that i¹ = 0, i² = i. This means that the entire current i bypasses R1 and flows through the short circuit R² = 0, the path of least resistance. Thus when a circuit is short circuited, as sho wn in Fig. 2.33(a), two things should be kept in mind:

• 1. The equivalent resistance Req = 0. [See what happens when R² = 0 in Eq. (2.37).]
• 2. The entire current flows through the short circuit.

As another e xtreme case, suppose R² = ∞, that is, R2 is an open circuit, as shown in Fig. 2.33(b). The current still flows through the path of least resistance, R1. By taking the limit of Eq. (2.37) as R² → ∞,we obtain Req = R1 in this case.

If we divide both the numerator and denominator by R1R2, Eq. (2.43) becomes

$$i_1 = \frac{G_1}{G_1 + G_2} i \tag{2.44a}$$ $$i_2 = \frac{G_2}{G_1 + G_2} i \tag{2.44b}$$

Thus, in general, if a current divider has N conductors (G1, G2, … , GN) in parallel with the source current i, the nth conductor (Gn) will have current

e current i, the nth conductor (

$$G_n$$

) will have current
\n

$$i_n = \frac{G_n}{G_1 + G_2 + \dots + G_N} i$$

\n(2.45)

Figure 2.33 (a) A shorted circuit, (b) an open circuit.

In general, it is often convenient and possible to combine resistors in series and parallel and reduce a resisti ve network to a single equivalent resistance Req. Such an equi valent resistance is the resistance between the designated terminals of the netw ork and must e xhibit the same i-v characteristics as the original network at the terminals.

Find Req for the circuit shown in Fig. 2.34. Example 2.9

Solution:

To get Req, we combine resistors in series and in parallel. The 6- Ω and 3-Ω resistors are in parallel, so their equivalent resistance is

$$6 \Omega \parallel 3\Omega = \frac{6 \times 3}{6 + 3} = 2 \Omega$$

(The symbol ∥ is used to indicate a parallel combination.) Also, the 1-Ω and 5-Ω resistors are in series; hence their equivalent resistance is

1 Ω + 5 Ω = 6 Ω

Thus the circuit in Fig. 2.34 is reduced to that in Fig. 2.35(a). In Fig. 2.35(a), we notice that the tw o 2-Ω resistors are in series, so the equivalent resistance is

$$2\Omega + 2\Omega = 4\Omega$$

This 4-Ω resistor is now in parallel with the 6-Ω resistor in Fig. 2.35(a); their equivalent resistance is

$$4 \Omega || 6 \Omega = \frac{4 \times 6}{4 + 6} = 2.4 \Omega$$

The circuit in Fig. 2.35(a) is now replaced with that in Fig. 2.35(b). In Fig. 2.35(b), the three resistors are in series. Hence, the equivalent resistance for the circuit is

$$R_{\text{eq}} = 4 \ \Omega + 2.4 \ \Omega + 8 \ \Omega = 14.4 \ \Omega$$

By combining the resistors in Fig. 2.36, find Req. Practice Problem 2.9

Answer: 11 Ω.

For Practice Prob. 2.9.

Calculate the equivalent resistance Rab Example 2.10 in the circuit in Fig. 2.37.

For Example 2.10.

Solution:

The 3-Ω and 6-Ω resistors are in parallel because they are connected to the same two nodes c and b. Their combined resistance is

$$3 \Omega \parallel 6 \Omega = \frac{3 \times 6}{3 + 6} = 2 \Omega$$

(2.10.1)

Similarly, the 12-Ω and 4-Ω resistors are in parallel since the y are connected to the same two nodes d and b. Hence

$$12 \Omega || 4 \Omega = \frac{12 \times 4}{12 + 4} = 3 \Omega$$

(2.10.2)

Also the 1-Ω and 5-Ω resistors are in series; hence, their equivalent re sistance is

$$1 \Omega + 5 \Omega = 6 \Omega \tag{2.10.3}$$

With these three combinations, we can replace the circuit in Fig. 2.37 with that in Fig. 2.38(a). In Fig. 2.38(a), 3-Ω in parallel with 6-Ω gives 2-Ω, as calculated in Eq. (2.10.1). This 2-Ω equivalent resistance is now in series with the 1-Ω resistance to give a combined resistance of 1 Ω +2Ω=3Ω. Thus, we replace the circuit in Fig. 2.38(a) with that in Fig. 2.38(b). In Fig. 2.38(b), we combine the 2-Ω and 3-Ω resistors in parallel to get

$$2 \Omega || 3 \Omega = \frac{2 \times 3}{2 + 3} = 1.2 \Omega$$

This 1.2-Ω resistor is in series with the 10-Ω resistor, so that

$$R_{ab} = 10 + 1.2 = 11.2 \ \Omega$$

Find Rab for the circuit in Fig. 2.39.

Answer: 19 Ω.

For Practice Prob. 2.10.

Find the equivalent conductance Geq for the circuit in Fig. 2.40(a).

Solution:

The 8-S and 12-S resistors are in parallel, so their conductance is

$$8 S + 12 S = 20 S$$

This 20-S resistor is now in series with 5 S as shown in Fig. 2.40(b) so that the combined conductance is

$$\frac{20 \times 5}{20 + 5} = 4$$

This is in parallel with the 6-S resistor. Hence,

$$G_{\text{eq}} = 6 + 4 = 10 \text{ S}$$

We should note that the circuit in Fig. 2.40(a) is the same as that in Fig. 2.40(c). While the resistors in Fig. 2.40(a) are expressed in siemens, those in Fig. 2.40(c) are expressed in ohms. To show that the circuits are the same, we find Req for the circuit in Fig. 2.40(c).

$$R_{\text{eq}} = \frac{1}{6} \left\| \left( \frac{1}{5} + \frac{1}{8} \right) \frac{1}{12} \right\| = \frac{1}{6} \left\| \left( \frac{1}{5} + \frac{1}{20} \right) \right\| = \frac{1}{6} \left\| \frac{1}{4} \right\|$$ $$= \frac{\frac{1}{6} \times \frac{1}{4}}{\frac{1}{6} + \frac{1}{4}} = \frac{1}{10} \Omega$$ $$G_{\text{eq}} = \frac{1}{R_{\text{eq}}} = 10 \text{ S}$$

This is the same as we obtained previously.

Calculate Geq in the circuit of Fig. 2.41. Practice Problem 2.11

Answer: 8 S.

For Practice Prob. 2.11.

Find io and vo in the circuit shown in Fig. 2.42(a). Calculate the power dissipated in the 3-Ω resistor.

Solution:

The 6-Ω and 3-Ω resistors are in parallel, so their combined resistance is

$$6 \Omega \parallel 3 \Omega = \frac{6 \times 3}{6 + 3} = 2 \Omega$$

Thus, our circuit reduces to that shown in Fig. 2.42(b). Notice that vo is not affected by the combination of the resistors because the resistors are

Example 2.11

Figure 2.40

For Example 2.11: (a) original circuit, (b) its equivalent circuit, (c) same circuit as in (a) but resistors are expressed in ohms.

Example 2.12

For Example 2.12: (a) original circuit, (b) its equivalent circuit.

in parallel and therefore have the same voltage vo. From Fig. 2.42(b), we can obtain vo in two ways. One way is to apply Ohm’s law to get

$$i = \frac{12}{4+2} = 2 \text{ A}$$

and hence, vo = 2i = 2 × 2 = 4 V. Another way is to apply voltage division, since the 12 V in Fig. 2.42(b) is divided between the 4-Ω and 2-Ω resistors. Hence,

$$v_o = \frac{2}{2+4} (12 \text{ V}) = 4 \text{ V}$$

Similarly, io can be obtained in two ways. One approach is to apply Ohm’s law to the 3-Ω resistor in Fig. 2.42(a) now that we know vo; thus,

$$v_o = 3i_o = 4 \qquad \Rightarrow \qquad i_o = \frac{4}{3} \text{ A}$$

Another approach is to apply current division to the circuit in Fig. 2.42(a) now that we know i, by writing

$$i_o = \frac{6}{6+3} i = \frac{2}{3} (2 \text{ A}) = \frac{4}{3} \text{ A}$$

The power dissipated in the 3-Ω resistor is

$$p_o = v_o i_o = 4 \left(\frac{4}{3}\right) = 5.333 \text{ W}$$

Find v1 and v2 in the circuit sho wn in Fig. 2.43. Also calculate i1 and i² and the power dissipated in the 12-Ω and 40-Ω resistors.

Answer: v¹ = 10 V, i1 = 833.3 mA, p1 = 8.333 W, v² = 20 V, i2 =500 mA, p² = 10 W.

Practice Problem 2.12

For Practice Prob. 2.12.

Figure 2.43

Example 2.13

For the circuit sho wn in Fig. 2.44(a), determine: (a) the v oltage vo, (b) the power supplied by the current source, (c) the power absorbed by each resistor.

Solution:

(a) The 6-k Ω and 12-k Ω resistors are in series so that their combined value is 6 + 12 = 18 kΩ. Thus the circuit in Fig. 2.44(a) reduces to that shown in Fig. 2.44(b). We now apply the current division technique to find i1 and i2.

$$i_1 = \frac{18,000}{9,000 + 18,000} (30 \text{ mA}) = 20 \text{ mA}$$ $$i_2 = \frac{9,000}{9,000 + 18,000} (30 \text{ mA}) = 10 \text{ mA}$$

Notice that the voltage across the 9-kΩ and 18-kΩ resistors is the same, and vo = 9,000i¹ = 18,000i² = 180 V, as expected. (b) Power supplied by the source is

$$p_o = v_o i_o = 180(30) \text{ mW} = 5.4 \text{ W}$$

(c) Power absorbed by the 12-kΩ resistor is

$$p = iv = i_2(i_2 R) = i_2^2 R = (10 \times 10^{-3})^2 (12,000) = 1.2 W$$

Power absorbed by the 6-kΩ resistor is

$$p = i_2^2 R = (10 \times 10^{-3})^2 (6,000) = 0.6 W$$

Power absorbed by the 9-kΩ resistor is

$$p = \frac{v_o^2}{R} = \frac{(180)^2}{9,000} = 3.6 \text{ W}$$

or

$$p = v_o i_1 = 180(20) \text{ mW} = 3.6 \text{ W}$$

Notice that the power supplied (5.4 W) equals the power absorbed (1.2 + 0.6 + 3.6 = 5.4 W). This is one way of checking results.

For the circuit shown in Fig. 2.45, find: (a) v1 and v2, (b) the power dis- Practice Problem 2.13 sipated in the 3-k Ω and 20-kΩ resistors, and (c) the power supplied by the current source.

Answer: (a) 135 V, 180 V, (b) 2.025 W, 540 mW, (c) 5.4 W.