Ch26Word - Chapter 26 p 1 CHAPTER 26 – DC Circuits 1 a For the current in the single loop we have I a = V R a r =(8.50 V(68.0 Ω 0.900 Ω =

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Unformatted text preview: Chapter 26 p. 1 CHAPTER 26 – DC Circuits 1. ( a ) For the current in the single loop, we have I a = V /( R a + r ) = (8.50 V)/(68.0 Ω + 0.900 Ω ) = 0.123 A. For the terminal voltage of the battery, we have V a = å – I a r = 8.50 V – (0.123 A)(0.900 Ω ) = 8.39 V . ( b ) For the current in the single loop, we have I b = V /( R b + r ) = (8.50 V)/(680 Ω + 0.900 Ω ) = 0.0125 A. For the terminal voltage of the battery, we have V b = å – I b r = 8.50 V – (0.0125 A)(0.900 Ω ) = 8.49 V . 2. The voltage across the bulb is the terminal voltage of the four cells: V = IR bulb = 4( å – Ir ); (0.62 A)(12 Ω ) = 4[2.0 V – (0.62 A) r ], which gives r = 0.23 Ω . 3. If we can ignore the resistance of the ammeter, for the single loop we have I = å / r ; 25 A = (1.5 V)/ r , which gives r = 0.060 Ω . 4. We find the internal resistance from V = å – Ir ; 9.8 V = [12.0 V – (60 A) r ], which gives r = 0.037 Ω . Because the terminal voltage is the voltage across the starter, we have V = IR ; 9.8 V = (60 A) R , which gives R = 0.16 Ω . 5. When the bulbs are connected in series, the equivalent resistance is R series = ? R i = 4 R bulb = 4(90 Ω ) = 360 Ω . When the bulbs are connected in parallel, we find the equivalent resistance from 1 / R parallel = ?(1/ R i ) = 4/ R bulb = 4/(90 Ω ), which gives R parallel = 23 Ω . 6. ( a ) When the bulbs are connected in series, the equivalent resistance is R series = ? R i = 3 R 1 + 3 R 2 = 3(40 Ω ) + 3(80 Ω ) = 360 Ω . ( b ) When the bulbs are connected in parallel, we find the equivalent resistance from 1 / R parallel = ?(1/ R i ) = (3/ R 1 ) + (3/ R 2 ) = [3/(40 Ω )] + [3/(80 Ω )], which gives R parallel = 8.9 Ω . 7. If we use them as single resistors, we have R 1 = 25 Ω ; R 2 = 70 Ω . When the resistors are connected in series, the equivalent resistance is R series = ? R i = R 1 + R 2 = 25 Ω + 70 Ω = 95 Ω . When the resistors are connected in parallel, we find the equivalent resistance from 1 / R parallel = ?(1/ R i ) = (1/ R 1 ) + (1/ R 2 ) = [1/(25 Ω )] + [1/(70 Ω )], which gives R parallel = 18 Ω . 8. Because resistance increases when resistors are connected in series, the maximum resistance is R series = R 1 + R 2 + R 3 = 500 Ω + 900 Ω + 1400 Ω = 2800 Ω = 2.80 k Ω . Because resistance decreases when resistors are connected in parallel, we find the minimum resistance from 1 / R parallel = (1/ R 1 ) + (1/ R 2 ) + (1/ R 3 ) = [1/(500 Ω )] + [1/(900 Ω )] + [1/(1400 Ω )], which gives R parallel = 261 Ω . Chapter 26 p. 2 9. The voltage is the same across resistors in parallel, but is less across a resistor in a series connection. We connect three 1.0- Ω resistors in series as shown in the diagram. Each resistor has the same current and thus the same voltage: V i = @ V = @ (6.0 V) = 2.0 V....
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch26Word - Chapter 26 p 1 CHAPTER 26 – DC Circuits 1 a For the current in the single loop we have I a = V R a r =(8.50 V(68.0 Ω 0.900 Ω =

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