Rnr rnc for the voltage-divider circuit figure with

78	C H A P T E R	T W O	r e s i s t i v e

in = in−1,		(2.62)

v0 + v1 + · · · vN = 0. (2.63)

Finally, Equations 2.60 through 2.63 can be solved to yield

i	1	V,	(2.64)
n =	R1 + R2 + · · · RN
		V,	(2.65)
		V,	(2.66)

This completes the analysis.

As was the case for the two-resistor voltage divider, the preceding analysis shows that series resistors divide voltage in proportion to their resistances. This follows from the Rn in the numerator of the right-hand side of Equation 2.66.

RS =V		(2.67)

FIGURE 2.34 The equivalence		R1
FIGURE 2.34 The equivalence
of series resistors.	. . .
of series resistors.

2.3 Circuit Analysis: Basic Method			C H A P T E R	T W O	79
	v o l t a g e - d i v i d e r c i r c u i t	Determine v1 and

2. Write the constituent relations

v0 = 3 V v1 = 10i1 v2 = 20i2.

	i1 − i2 = 0.	(2.71)
		(2.72)

Now eliminate i1 and i2 from Equations 2.69, 2.70, and 2.71, to obtain

v1 =v2	.	(2.73)

	+	−3 V +v2 2				10 Ω	+	i2
		−3 V +v2 2			i1	10 Ω
		R1		+		v1

				-
		v(t)	R2	+			v2	20 Ω
				+			v2
generator				v2	-		-
generator	-			v2	-		-