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1Bch26 - Chapter 26 There are two main types of resistor...

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Chapter 26 There are two main types of resistor configurations: series and parallel. For each of these, here’s how the currents and voltages across each resistor are related, and how to compute the equivalent resistance. Resistors in series: Suppose we have resistors R 1 , R 2 , ........... , R N connected in series, i.e., strung together sequentially in the same branch of a circuit. Then the current through each resistor is the same: I 1 = I 2 = ...... = I N = I Furthermore, the total voltage across the combination is the sum of the voltage V j across each resistor R j . That is, V = N X j =1 V j = N X j =1 I j R j = I N X j =1 R j = IR eq The equivalent resistance of the series combination is therefore R eq = N X j =1 R j Resistors in parallel: Suppose we have resistors R 1 , R 2 , ....... , R N connected in parallel, so that each resistor occupies its own branch. Then the voltage across each resistor is the same: V 1 = V 2 = .... = V N = V Furthermore, the total current passing through the combination is the sum of the current I j passing through each resistor R j . That is, I = N X j =1 I j = N X j =1 V j R j = V N X j =1 1 R j = V R eq The equivalent resistance of the parallel combination is therefore given by 1 R eq = N X j =1 1 R j Problem 1: Five resistors are assembled as shown in the figure below. The voltage in the battery is V ab . 1
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1: Find the equivalent resistance of the circuit. 2: Find the current and voltage across each resistor. °± ² ° ± ° ³ ± ³ ² ³ ³ ³ ´ ³ ° ´ ´ ´ ² ´ ³ ´ ± ´ ° ± Solution: This problem is identical to problem 1 of 1Bch24.pdf, except with the capacitors replaced by resistors. We solve it in very much the same way, starting with part 1 , which asks to find the equivalent resistance. Resistors R 3 and R 4 are connected in series. Therefore, their equivalent resistance R 34 is R 34 = R 3 + R 4 Equivalent resistor R 34 is connected in parallel with resistor R 2 . Therefore, their equivalent resistance R 234 is given by 1 R 234 = 1 R 2 + 1 R 34 R 234 = R 2 R 34 R 2 + R 34 Finally, equivalent resistor R 234 is connected in series with resistor R 1 and resistor R 5 . Therefore, their equivalent resistance R eq is R eq = R 1 + R 234 + R 5 This is the equivalent resistance of the circuit. We now proceed to part 2 , where we reverse the steps in part 1. We start with a single resistor with resistance R eq and progressively “decompose” the circuit, alternately solving 2
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for the current and voltage at each stage of “decomposition”. We begin by finding the total current I through the circuit, given by I = V ab R eq We now notice that because R 1 , R 234 , and R 5 , are connected in series, the current through each is the same, and is equal to I : I 1 = I 234 = I 5 = I = V ab R eq The voltage across each of these resistors is V 1 = I 1 R 1 = V ab R 1 R eq V 5 = I 5 R 5 = V ab R 5 R eq V 234 = I 234 R 234 = V ab R 234 R eq Because R 2 and R 34 are connected in parallel, the voltage across each is the same, and is equal to V 234 . That is, V 2 = V 34 = V 234 = V ab R 234 R eq The currents through resistors R 2 and R 34 are I 2 = V 2 R 2 = V ab R 234 R 2 R eq I 34 = V 34 R 34 = V ab R 234 R 34 R eq Because resistors R 3 and R 4
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