Physics Solution Manual for 1100 and 2101

The sum of the potential rises must equal the sum of

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Unformatted text preview: ages across them are equal. Thus, we can also use Ohm's law to find the current through the 16.0-Ω resistor. The currents that flow through the 8.00-Ω and the 16.0-Ω resistors combine to give the total current that flows through the 20.0-Ω resistor. Similar reasoning can be used to find the current through the 9.00-Ω resistor. SOLUTION a. The voltage across the 8.00-Ω resistor is V8 = (0.500 A)(8.00 Ω) = 4.00 V . Since this is also the voltage that is across the 16.0-Ω resistor, we find that the current through the 16.0-Ω resistor is I16 = (4.00 V)/(16.0 Ω ) = 0.250 A . Therefore, the total current that flows through the 20.0-Ω resistor is I20 = 0.500 A + 0.250 A = 0.750 A b. The 8.00-Ω and the 16.0-Ω resistors are in parallel, so their equivalent resistance can be 1 1 1 1 obtained from Equation 20.17, = + + + ... , and is equal to 5.33 Ω. Therefore, Rp R1 R2 R3 the equivalent resistance of the upper branch of the circuit is Rupper = 5.33 Ω + 20.0 Ω = 25.3 Ω , since the 5.33-Ω resistance is in series with the 20.0-Ω 1100 ELECTRIC CIRCUITS resistance. Using Ohm's law, we find that the voltage across the upper branch must be V = (0.750 A)(25.3 Ω ) = 19.0 V . Since the lower branch is in parallel with the upper branch, the voltage across both branches must be the same. Therefore, the current through the 9.00-Ω resistor is, from Ohm's law, Vlower 19.0 V = 2.11 A R9 9.00 Ω ______________________________________________________________________________ I9 = = 72. REASONING The resistance R of each of the identical resistors determines the equivalent resistance Req of the entire circuit, both in its initial form (six resistors) and in its final form (five resistors). We will calculate the initial and final equivalent circuit resistances by replacing groups of resistors that are connected either in series or parallel. The resistances of the replacement resistors are found either from RS = R1 + R2 + R3 + L (Equation 20.16) for resistors connected in series, or from 1 1 1 1 =+ + + L (Equation 20.17), for resistors RP R1 R2 R3 connected in parallel. Once the equivalent resistance Req of the circuit is determined, the current I supplied by the V battery (voltage = V) is found from I = (Equation 20.2). We are given the decrease ∆I Req in the battery current, which is equal to the final current If minus the initial current I0: ∆I = I f − I 0 = −1.9 A (1) The algebraic sign of ∆I is negative because the final current is smaller than the initial current. SOLUTION Beginning with the circuit in its initial form, we see that resistors 1, 3, and 5 are connected in series (see the drawing). These three resistors, according to Equation 20.16, may be replaced with a single resistor RS that has three times the resistance R of a single resistor: R1 R2 R4 R3 R6 R5 R S = 3R R2 The resistor RS is connected in parallel with the resistor R4 (see the drawing), so these two resistors may be replaced by an equivalent resistor RP found from Equation 20.17: 1 RP = R P −1 1 1 = + R S R4 −1 1 1 = + 3R R −1 3 1 = + 3 R 3R R4 RS = 3 R R6 −1 4 = 3R −1 = 3R 4 Chapter 20 Problems After making this replacement, the three remaining resistors (R2, RP, and R6) are connected in series across the battery (see the drawing). The initial equivalent resistance Req,0 of the entire circuit, then, is found from Equation 20.16: R eq,0 = RP + R2 + R4 = R2 RP = 3R/4 (2) R1 R2 R5 R6 R3 R eq,f = 5 R From I = R6 3R 11R + 2R = 4 4 Next, we consider the circuit after the resistor R4 has been removed. The remaining five resistors are connected in series (see the drawing). From Equation 20.16, then, the final equivalent resistance Req,f of the entire circuit is five times the resistance R of a single resistor: 1101 (3) V (Equation 20.2), the initial and final battery currents are Req I0 = V Req,0 and If = V Req,f (4) Substituting Equations (2), (3), and (4) into Equation (1), we obtain ∆I = I f − I 0 = V V V V V 1 4 9V − = − = − =− Req,f Req,0 5R 11R R 5 11 55R 4 (5) Solving Equation (5) for R yields R=− 73. 9 ( 35 V ) 9V =− = 3.0 Ω 55∆I 55 ( −1.9 A ) SSM REASONING The terminal voltage of the battery is given by Vterminal = Emf – Ir , where r is the internal resistance of the battery. Since the terminal voltage is observed to be one-half of the emf of the battery, we have Vterminal = Emf/2 and I = Emf / ( 2r ) . From Ohm's law, the equivalent resistance of the circuit is R = emf / I = 2 r . We can also find the equivalent resistance of the circuit by considering that the identical bulbs are in parallel across the battery terminals, so that the equivalent resistance of the N bulbs is found from 1102 ELECTRIC CIRCUITS 1 N = R p Rbulb Rp = or Rbulb N This equivalent resistance is in series with the battery, so we find that the equivalent resistance of the circuit is R bulb R = 2r = +r N This expression can be solved for N. SOLUTION Solving the above expression for N, we have 15 Ω = 30 2r − r r 0.50 Ω ______________________________________________________________________________ N= R bulb = R...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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