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74 Direct-Current Circuit Analysis
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Using the first method, first change the resistances Rn into conductances Gn. This gives G1 = 1/R1 = 1/22 = 0.04545 S, G2 = 1/R2 = 1/47 = 0.02128 S, and G3 = 1/R3 = 1/68 = 0.01471 S. Adding these gives G = 0.08144 S. 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. 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 = 0.5455 A, I2 = E/R2 = 12/47 = 0.2553 A, and I3 = E/R3 = 12/68 = 0.1765 A. Adding these gives I = I1 + I2 + I3 = 0.5455 + 0.2553 + 0.1765 = 0.9773 A, which rounds off to 0.98 A.
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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 dissipated by any one of the resistances Rn, based on the formula Pn = I 2Rn.
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Problem 5-7 Suppose we have a series circuit with a supply of 150 V and three resistances: R1 = 330 , R2 = 680 , and R3 = 910 . What is the power dissipated by R2 First, find the current that flows through the circuit. Calculate the total resistance first. Because the resistors are in series, the total is R = 330 + 680 + 910 = 1920 . The current is I = 150/1920 = 0.07813 A. The power dissipated by R2 is therefore P2 = I 2R2 = 0.07813 0.07813 680 = 4.151 W. Round this off to three significant digits, because that s all we have in the data, to obtain 4.15 W. The total wattage dissipated in a series circuit is equal to the sum of the wattages dissipated in each resistance. Problem 5-8 Calculate the total dissipated power P in the circuit of Problem 5-7 by two different methods. First, let s figure out the power dissipated by each of the three resistances separately, and then add the figures up. The power P2 is already known. Let s use all the significant digits we have while we calculate. Thus, as found in Problem 5-7, P2 = 4.151 W. Recall that the current is I = 0.07813 A. Then P1 = 0.07813 0.07813 330 = 2.014 W, and P3 = 0.07813 0.07813 910 = 5.555 W. Adding the three power figures gives us P = P1 + P2 + P3 = 2.014 + 4.151 + 5.555 = 11.720 W. We should round this off to 11.7 W. The second method is to find the total series resistance and then calculate the power. The series resistance is R = 1920 , as found in Problem 5-7. Then P = I 2R = 0.07813 0.07813 1920 = 11.72 W. Again, we should round this to 11.7 W.
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When resistances are wired in parallel, they each consume power according to the same formula, P = I 2R. But the current is not the same in each resistance. An easier method to find the power Pn dissipated by each of the various resistances Rn is to use the formula Pn = E 2/Rn, where E is the voltage of the supply or battery. This voltage is the same across every branch resistance in a parallel circuit.
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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 = 3.0 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 = 3.0 3.0 = 9.0. Then the wattages dissipated by resistances R1, R2, and R3 respectively are P1 = 9.0/22 = 0.4091 W, P2 = 9.0/47 = 0.1915 W, and P3 = 9.0/68 = 0.1324 W. These should be rounded off to P1 = 0.41 W, P2 = 0.19 W, and P3 = 0.13 W. (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 = 0.4091 + 0.1915 + 0.1324 = 0.7330 W. Now that we ve finished the calculation, we should round it off to 0.73 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 = 12.28 . Then P = E 2/R = 9.0/12.28 = 0.7329 W. Now that the calculation is done, this can be rounded to 0.73 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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The current in a series circuit is the same at every point along the way. The voltage across any resistance in a parallel combination of resistances is the same as the voltage across any other resistance, or across the whole set of resistances. The voltages across resistances in a series circuit always add up to the supply voltage. The currents through resistances in a parallel circuit always add up to the total current drawn from the supply. The total wattage consumed in a series or parallel circuit is always equal to the sum of the wattages dissipated in each of the resistances. Now, let s get acquainted with two of the most famous laws that govern dc circuits. These rules are broad and sweeping, and they make it possible to analyze complicated series-parallel dc networks.
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