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.

Two separate capacitors, C1 and C2 C1 = 36 micro-Coulomb on 3 micro-Farad   C2 = 72 uC on X uF, X = 3.5 , if zero X =2   C2 had a gap of 0.2m maintained by a compressed plastic spring inside the gap, the natural spring length was 0.5m, the compressed spring length was 0.2 m. Spring constant = 8,000 micro-Newton/ meter Action: Connected the two capacitors in parallel Part A Find Q2-new, C2-new, new gap   Part B Find the initial total energy, the final total energy

Suppose you are given several capacitors whose capacitances are within the range 12 to 50 nF.You connect all the capacitors in series and let your three partners measure the equivalent capacitance. Three different measurements have been obtained: 8, 44 and 102 nF. Which ofthese three you would assume to be correct?

(d) Two capacitors are in series across a 60V source. If one is a 1 µF capacitor with a 15 V across it, what is the capacitance of the other capacitor?