# ass16 - SOLUTIONS A.2 IDENTIFY:It may appear that the meter...

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SOLUTIONS, ASSIGNMENT 16 26.2. I DENTIFY : It may appear that the meter measures X directly. But note that X is in parallel with three other resistors, so the meter measures the equivalent parallel resistance between ab . S ET U P : We use the formula for resistors in parallel. E XECUTE : 1/(2.00 Ω) = 1/ X + 1/(15.0 Ω) + 1/(5.0 Ω) + 1/(10.0 Ω), so X = 7.5 Ω. E VALUATE : X is greater than the equivalent parallel resistance of 2.00 Ω. 26.6. I DENTIFY : The potential drop is the same across the resistors in parallel, and the current into the parallel combination is the same as the current through the 45.0-Ω resistor. (a) S ET U P : Apply Ohm’s law in the parallel branch to find the current through the 45.0-Ω resistor. Then apply Ohm’s law to the 45.0-Ω resistor to find the potential drop across it. E XECUTE : The potential drop across the 25.0-Ω resistor is V 25 = (25.0 Ω)(1.25 A) = 31.25 V. The potential drop across each of the parallel branches is 31.25 V. For the 15.0-Ω resistor: I 15 = (31.25 V)/(15.0 Ω) = 2.083 A. The resistance of the 10.0- Ω + 15.0 Ω combination is 25.0 Ω, so the current through it must be the same as the current through the upper 25.0 Ω resistor: I 10+15 = 1.25 A. The sum of currents in the parallel branch will be the current through the 45.0-Ω resistor. I Total = 1.25 A + 2.083 A + 1.25 A = 4.58 A Apply Ohm’s law to the 45.0 Ω resistor: V 45 = (4.58 A)(45.0 Ω) = 206 V (b) S ET U P : First find the equivalent resistance of the circuit and then apply Ohm’s law to it. E XECUTE : The resistance of the parallel branch is 1/ R = 1/(25.0 Ω) + 1/(15.0 Ω) + 1/(25.0 Ω), so R = 6.82 Ω. The equivalent resistance of the circuit is 6.82 Ω + 45.0 Ω + 35.00 Ω = 86.82 Ω. Ohm’s law gives V Bat = (86.62 Ω)(4.58 A) = 398 V. E VALUATE : The emf of the battery is the sum of the potential drops across each of the three segments (parallel branch and two series resistors). 26.11. I DENTIFY : For resistors in parallel, the voltages are the same and the currents add. eq 1 2 1 1 1 R R R = + so 1 2 eq 1 2 , R R R R R = + For resistors in series, the currents are the same and the voltages add. eq

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## This note was uploaded on 03/11/2010 for the course PHYSICS 1523 taught by Professor Dfs during the Fall '10 term at Dallas.

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ass16 - SOLUTIONS A.2 IDENTIFY:It may appear that the meter...

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