Chap_21b_soln[1]

Chap_21b_soln[1] - 21-53: Picture the Problem: Six...

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21-53: Picture the Problem : Six resistors are connected in the manner indicated in the diagram, and a 9.0-V battery is connected to terminals A and B. Strategy: Find the equivalent resistance of the right-hand portion of this circuit by combining resistors 4, 5, and 6 in parallel, and then combining that group in series with resistor 3. The voltage drops across resistors 1, 2, and are each 9.0 V. Ohm’s Law can then be used to find eq R eq R 3 d . 12 , , an I II Use Ohm’s Law together with 3 I to find the voltage drop across resistor 3, then subtract that result from 9.0 V to find the voltage drop across resistors 4, 5, and 6. Ohm’s Law can then be applied to each of those resistors to find 45 , , 6 and . I Solution: 1. (a) Use Ohm’s Law to find 1 : I 1 9.0 V 6.0 A 1.5 I  2. Repeat step 1 to find 2 : I 2 9.0 V 3.6 A 2.5 I 3. Find for the right-hand portion of the circuit: eq R 1 eq 111 6.3 7.9 4.8 3.3 8.1 R         4. Apply Ohm’s Law to the right-hand branch: 3 9.0 V 1.14 A 1.1 A 7.9 I 5. Find across resistors 4, 5, and 6: V  33 9.0 V 1.14 A 6.3 1.8 V VI R   6. Apply Ohm’s Law to find 4 : I 4 1.8 V 0.38 A 4.8 I 7. Repeat step 6 to find 5 : I 5 1.8 V 0.55 A 3.3 I 8. Repeat step 6 to find 6 : I 6 1.8 V 0.22 A 8.1 I 9. (b) There is a potential drop across resistors 4, 5, and 6 (1.8 V in this case), resulting in a potential difference across the 6.3- resistor that is less than the 9.0 V that is across the 1.5- resistor.
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Chap_21b_soln[1] - 21-53: Picture the Problem: Six...

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