EXAMPLE 19-3 Series and parallel resistors. Two 100-0 resistors are con- nected (a) in parallel, and (b) in series, to a 24.0-V battery (Fig. 19–6). What is the current through each resistor and what is the equivalent resistance of each circuit? APPROACH We use Ohm's law and the ideas just discussed for series and parallel connections to get the current in each case. We can also use Eqs. 19–3 and 19-4. SOLUTION (a) Any given charge (or electron) can flow through only one or the other of the two resistors in Fig. 19-6a. Just as a river may break into two streams when going around an island, here too the total current I from the battery (Fig. 19–6a) splits to flow through each resistor, so I equals the sum of the separate currents through the two resistors: I = 1, + 4. |The potential difference across each resistor is the battery voltage V = 24.0 V. Applying Ohm's law to each resistor gives V V 24.0 V, 24.0 V I = 1, + I, R 0.24 A + 0.24 A = 0.48 A. R2 100 Ω 100 N |The equivalent resistance is 24.0 V Reg 50 Ω. 0.48 A We could also have obtained this result from Eq. 19–4: 1 1 2 1 Rea 100Ω, 100 η 50 N 100 Ω so Re = 50 2. (b) All the current that flows out of the battery passes first through R, and then through R, because they lie along a single path, Fig. 19–6b. So the current I is the same in both resistors; the potential difference V across the battery equals the total change in potential across the two resistors: V = V, + V½. Ohm's law gives V = IR, + IR, = I(R, + R.). Hence V 24.0 V = 0.120 A. R + R2 100 Ω + 100 Ω The equivalent resistance, using Eq. 19–3, is Reg = R1 + R2 = 200 0. We can also get Req by thinking from the point of view of the battery: the total resistance Reg must equal the battery voltage divided by the current it delivers: V 24.0 V Reg 0.120 A 200 Ω. V = 24.0 V R1 R2 ww (a) V = 24.0 V ww ww R1 R2 (b) FIGURE 19-6 Example 19–3.

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(II) Design a “voltage divider” (see Example 19–3) that
would provide one-fifth (0.20) of the battery voltage across R₂
Fig. 19–6. What is the ratio R₁/R₂

EXAMPLE 19-3 Series and parallel resistors. Two 100-0 resistors are con-
nected (a) in parallel, and (b) in series, to a 24.0-V battery (Fig. 19–6). What is the
current through each resistor and what is the equivalent resistance of each circuit?
APPROACH We use Ohm's law and the ideas just discussed for series and parallel
connections to get the current in each case. We can also use Eqs. 19–3 and 19-4.
SOLUTION (a) Any given charge (or electron) can flow through only one or
the other of the two resistors in Fig. 19-6a. Just as a river may break into two
streams when going around an island, here too the total current I from the
battery (Fig. 19–6a) splits to flow through each resistor, so I equals the sum of
the separate currents through the two resistors:
I = 1, + 4.
|The potential difference across each resistor is the battery voltage V = 24.0 V.
Applying Ohm's law to each resistor gives
V
V
24.0 V, 24.0 V
I = 1, + I,
R
0.24 A + 0.24 A = 0.48 A.
R2
100 Ω
100 N
|The equivalent resistance is
24.0 V
Reg
50 Ω.
0.48 A
We could also have obtained this result from Eq. 19–4:
1
1
2
1
Rea
100Ω, 100 η
50 N
100 Ω
so Re = 50 2.
(b) All the current that flows out of the battery passes first through R, and then
through R, because they lie along a single path, Fig. 19–6b. So the current I is
the same in both resistors; the potential difference V across the battery equals
the total change in potential across the two resistors:
V = V, + V½.
Ohm's law gives V = IR, + IR, = I(R, + R.). Hence
V
24.0 V
= 0.120 A.
R + R2
100 Ω + 100 Ω
The equivalent resistance, using Eq. 19–3, is Reg = R1 + R2 = 200 0. We
can also get Req by thinking from the point of view of the battery: the total
resistance Reg must equal the battery voltage divided by the current it delivers:
V
24.0 V
Reg
0.120 A
200 Ω.

V = 24.0 V
R1
R2
ww
(a)
V = 24.0 V
ww
ww
R1
R2
(b)
FIGURE 19-6 Example 19–3.

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