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CourseNotes.32 - resistor On the way to V 2 we pass through...

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Figure 26: The circuit for Problem 27.33 (Note that I will generally use P to denote a simplified resistor instead of R eq .) By simplifying these resistors, we now have a single loop circuit with 4 resistors in series and two batteries. The next step is to figure out what the current is in this single loop circuit. Since the rest of the resistors are in series, we can simplify them to a single resistor of resistance P 2 = 3 R + R 2 = 7 R 2 with two voltage sources. Hence, the current flowing through the circuit is i = V R = E 2 - E 1 P 2 Now that we know the current flowing through the circuit, we can take one step back to our first simplified circuit and start traversing the circuit to find the voltages V 2 and V 2 . Since the top right corner of the circuit is grounded (set at 0 potential) we will start there and work our way counter-clockwise around the circuit. We must be careful here because the current will be flowing with the direction which we are going and hence, we will lose potential as we go through each
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Unformatted text preview: resistor. On the way to V 2 we pass through two resistors and E 1 . V 2 = 0-iR-iR- E 1 =-E 1-2 E 2- E 1 P 2 R =-E 1-4 7 ( E 2- E 1 ) =-4 7 E 2-3 7 E 1 =-9 V Finding V 1 is now a simple matter of going one resistor further. V 1 = V 2-iR =-4 7 E 2-3 7 E 1-2 7 ( E 2- E 1 ) =-6 7 E 2-1 7 E 1 =-11 V Problem 27.35 In figure 27, R 1 = 2Ω, R 2 = 5Ω, and the battery is ideal. What value of R 3 maximizes the dissipation rate in resistance 3? Figure 27: The circuit for Problem 27.35 The power dissipated by resistor three is given by: P 3 = i 3 V 3 . Hence, our first job is to find the current flowing through and the potential difference across resistor three. Lets start with the potential difference. If we simplify R 2 and R 3 down to an equivalent resistor we find that their equivalent resistance is P = R 2 R 3 R 2 + R 3 32...
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