If r1 r2 rn r then r 239 n for example if four

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . The equivalent resistance is 1 1 1 1 = + + ··· + Req R1 R2 RN (2.38) Note that Req is always smaller than the resistance of the smallest resistor in the parallel combination. If R1 = R2 = · · · = RN = R , then R (2.39) N For example, if four 100- resistors are connected in parallel, their equivalent resistance is 25 . 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 Req = Geq = G1 + G2 + G3 + · · · + GN (2.40) Conductances in parallel behave as a single conductance whose value is equal to the sum of the individual conductances. | v v where Geq = 1/Req , G1 = 1/R1 , G2 = 1/R2 , G3 = 1/R3 , . . . , GN = 1/RN . Equation (2.40) states: | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents 44 PART 1 DC Circuits The equivalent conductance of resistors connected in parallel is the sum of their individual conductances. i v a + − v Req or Geq b Figure 2.32 Equivalent circuit to Fig. 2.31. 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 way as the equivalent resistance of series resistors. In the same manner, the equivalent conductance of resistors in series is obtained just the same 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 1 1 1 1 1 = + + + ··· + Geq G1 G2 G3 GN (2.41) Given the total current i entering node a in Fig. 2.31, how do we obtain current i1 and i2 ? We know that the equivalent resistor has the same voltage, or iR1 R2 (2.42) v = iReq = R1 + R 2 Combining Eqs. (2.33) and (2.42) results in i1 = i i2 = i i1 = 0 R1 R2 = 0 i i2 = 0 i1 = i R1 R2 = ∞ (b) v v | (a) A shorted circuit, (b) an open circuit. | i2 = R1 i R1 + R 2 (2.43) which shows that the to...
View Full Document

Ask a homework question - tutors are online