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physics solution

# physics solution - SLTNS whw 3 CH 21 AND MPTWO Problem#s...

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SLTNS whw # 3 CH 21 AND MPTWO Problem #s: 45,51,77,78,79,81,91,93 45. Picture the Problem : Four resistors and a switch are connected to a battery as shown in the diagram at the right. Strategy: The symmetric nature of the circuit gives clues to its behavior. Because the total resistances for the two paths on either side of the switch are the same, equal currents will flow through each branch. Because the top two resistors in each branch are the same, the voltage drops across each will be the same as well. That means the electric potential at either side of the switch will be the same, and therefore the potential difference across the switch is zero and no current will flow through the switch when it is closed. Use Ohm’s Law and the rules for calculating equivalent resistance (equations 21-7 and 21-10) to verify these observations. Solution: 1. (a) Because the resistors are identical, the potential is the same at both sides of the switch regardless of whether it is open or closed. Therefore the current through the battery will stay the same when the switch is closed. 2. (b) Find the current 0 I through the battery when the switch is open: The circuit is two parallel branches of resistors connected in series. 0 eq 1 2 I R R R R 3. Find the current I through the battery when the switch is closed. The circuit is a series connection of two sets of resistors that are connected in parallel. 0 1 1 1 1 1 1 eq R R R R I I R R Insight: If the bottom two resistors were each doubled to 2 R , the current 0 I would decrease but the current I when the switch is closed would still equal 0 I . 51. Picture the Problem : Four resistors and two batteries are connected as shown in the circuit diagram at right. Strategy: The circuit can be analyzed by applying Kirchoff’s rules. First apply the Junction Rule to point A in the circuit, then apply the Loop Rule to two loops, the left-hand loop 1 and the outside loop 2 labeled in the diagram. These three equations can be combined to algebraically find 1 2 3 , , and . I I I From the currents we can find the potential difference between the points A and B. Solution: 1. (a) Apply the Junction Rule to point A: 1 2 3 (i) I I I K K 2. Apply the Loop Rule to loop 1, beginning in lower left-hand corner, and solve for 3 I : 1 1 3 3 1 4 1 4 3 1 1 3 3 0 12 V 12 V 13.7 10 A (ii) 1.2 I R I R I R R R I I I R R K K 3. Apply the Loop Rule to loop 2, beginning in lower left-hand corner, and solve for 2 I : 1 1 2 2 1 4 1 4 2 1 1 2 2 0 12 V 9.0 V 12.0 9.0 V 3.0 V 13.7 (iii) 6.7 6.7 I R I R I R R R I I I R R K K

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4. Substitute equations (ii) and (iii) into equation (i).
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