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What is the total current drawn by the circuit described in problem 5-5 There are two ways to go at this. One method involves finding the total resistance, R, of R1, R2, and R3 in parallel, and then calculating I based on R. Another, perhaps easier, way is to find the currents through R1, R2, and R3 individually, and then add them up. Using the first method, first change the resistances Rn into conductances Gn. This gives G1 1/R1 1/22 0.04545 siemens, G2 1/R2 1/47 0.02128 siemens, and G3 1/R3 1/68 0.01471 siemens. Adding these gives G 0.08144 siemens. The resistance is therefore R 1/G 1/0.08144 12.279 . Use Ohm s Law to find I E/R 12/12.279 0.98 A 980 mA. Note that extra digits are used throughout the calculation, rounding off only at the end. Now let s try the other method. Find I1 E/R1 12/22 0.5455A, 12 E/R2 12/47 0.2553 A, and 13 E/R3 12/68 0.1765 A. Adding these gives I I1 I2 3 0. 5455 0.2553 0.1765 0.9773 A, rounded off to 0.98 A. Allowing extra digits during the calculation saved my having to explain away a mathematical artifact. It could save you similar chagrin some day. Doing the problem both ways helped me to be sure I didn t make any mistakes in finding the answer to this problem. It could have the same benefit for you, when the option presents itself.

Problem 5-9 Suppose a dc circuit contains three resistances R1 = 22 , R2 = 47 , and R3 = 68 across a battery that supplies a voltage of E = 30 V Find the power dissipated by each resistance Let s find the square of the supply voltage, E 2, first We ll be needing this figure often: E 2 = 30 30 = 90 Then the wattages dissipated by resistances R1, R2, and R3 respectively are P1 = 90/22 = 04091 W, P2 = 90/47 = 01915 W, and P3 = 90/68 = 01324 W These should be rounded off to P1 = 041 W, P2 = 019 W, and P3 = 013 W.

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Let s switch back now to series circuits. This is a good exercise: getting used to thinking in different ways and to changing over quickly and often. When calculating the power in a circuit containing resistors in series, all you need to do is find out the current, I, that the circuit is carrying. Then it s easy to calculate the power Pn, based on the formula Pn I 2Rn.

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(But let s remember the values to four significant figures for the next problem!) In a parallel circuit, the total dissipated wattage is equal to the sum of the wattages dissipated by the individual resistances Problem 5-10 Find the total consumed power of the resistor circuit in Problem 5-9 using two different methods The first method involves adding P1, P2, and P3 Let s use the four-significant-digit values to avoid the possibility of encountering the rounding-off bug The total power thus calculated is P = 04091 + 01915 + 01324 = 07330 W Now that we ve finished the calculation, we should round it off to 073 W The second method involves finding the net resistance R of the parallel combination You can do this calculation yourself Determining it to four significant digits, you should get a net resistance of R = 1228 .

Problem 5-7

Then P = E 2/R = 90/1228 = 07329 W Now that the calculation is done, this can be rounded to 073 W It s the Law! In electricity and electronics, dc circuit analysis can be made easier if you are acquainted with certain axioms, or laws Here they are:.

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