Physics Solution Manual for 1100 and 2101

# This expression can be solved for the speed v

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Unformatted text preview: SOLUTION Ohm's law (Equation 20.2), V = IR , gives the result directly: V 9.0 V R= = = 82 Ω I 0.11 A ______________________________________________________________________________ 112. REASONING AND SOLUTION Label the currents with the resistor values. Take I3 to the right, I2 to the left and I1 to the right. Applying the loop rule to the top loop (suppressing the units) gives I1 + 2.0 I2 = 1.0 (1) and to the bottom loop gives 2.0 I2 + 3.0 I3 = 5.0 (2) Applying the junction rule to the left junction gives I2 = I1 + I3 (3) Solving Equations (1) , (2) and (3) simultaneously, we find I2 = 0.73 A . The positive sign shows that the assumed direction is correct. That is, to the left . ______________________________________________________________________________ 113. REASONING AND SOLUTION Solving Equation 20.5 for α yields 43.7 Ω –1 38.0 Ω –1 α= = = 0.0050 ( C° ) T – T0 55 °C – 25 °C ______________________________________________________________________________ R –1 R0 1126 ELECTRIC CIRCUITS 114. REASONING The magnitude q of the charge on one plate of a capacitor is given by Equation 19.8 as q = CV1, where C = 9.0 µ F and V1 is the voltage across the capacitor. Since the capacitor and the resistor R1 are in parallel, the voltage across the capacitor is equal to the voltage across R1. From Equation 20.2 we know that the voltage across the 4.0-Ω resistor is given by V1 = IR1, where I is the current in the circuit. Thus, the charge can be expressed as q = CV1 = C ( I R1 ) The current is equal to the battery voltage V divided by the equivalent resistance of the two resistors in series, so that V V I= = RS R1 + R2 Substituting this result for I into the equation for q yields V q = C ( I R1 ) = C R + R R1 2 1 SOLUTION The magnitude of the charge on one of the plates is V 12 V −6 −5 q = C R1 = 9.0 × 10 F ( 4.0 Ω ) = 7.2 × 10 C 4.0 Ω + 2.0 Ω R1 + R2 ______________________________________________________________________________ ( ) 115. REASONING As discussed in Section 20.11, some of the current (6.20 mA) goes directly through the galvanometer and the remainder I goes through the shunt resistor. Since the resistance of the coil RC and the shunt resistor R are in parallel, the voltage across each is the same. We will use this fact to determine how much current goes through the shunt resistor. This value, plus the 6.20 mA that goes through the galvanometer, is the maximum current that this ammeter can read. SOLUTION The voltage across the coil resistance is equal to the voltage across the shunt resistor, so 6.20 × 10 A ) ( 20.0 Ω I ) ( 24.8 × 10 Ω (1444 24444 ) = (144 2444) 4 3 4 3 −3 Voltage across coil resistance −3 Voltage across shunt resistor So I = 5.00 A. The maximum current is 5.00 A + 6.20 mA = 5.01 A . ______________________________________________________________________________ Chapter 20 Problems 1127 116. REASONING The power Pn delivered to any one of the three resistors is equal to the 2 2 product of the current squared (I ) and the resistance Rn, or Pn = I Rn, where n = 1, 2, or 3. In each case, the resistance is known, and Ohm’s law can be used to find the current. Ohm’s law states that the current in the circuit (which is also the current through each of the resistors) equals the voltage V of the battery divided by the equivalent resistance RS of the three resistors: I = V/RS. Since the resistors are connected in series, we can obtain the equivalent resistance by adding the three resistances. SOLUTION The power Pn supplied to any one of the three resistors is Pn = I 2 Rn (n = 1, 2, or 3) (20.6b) The current I depends on the voltage V of the battery and the equivalent resistance RS of the three resistors through Ohm’s law: I= V RS (20.2) Substituting Equation 20.2 into Equation 20.6b gives 2 V Pn = I 2 Rn = Rn RS (n = 1, 2, or 3) (1) Since the three resistors are wired in series, the equivalent resistance RS is the sum of the resistances: RS = R1 + R2 + R3 (Equation 20.16). Substituting this relation into Equation (1) yields 2 2 V V Pn = Rn = Rn RS R1 + R2 + R3 (n = 1, 2, or 3) The power delivered to each resistor is: 2 2 V 24 V P= 1 R1 = 2.0 Ω + 4.0 Ω + 6.0 Ω ( 2.0 Ω ) = 8.0 W R1 + R2 + R3 2 2 2 2 V 24 V P2 = R2 = 2.0 Ω + 4.0 Ω + 6.0 Ω ( 4.0 Ω ) = 16 W R1 + R2 + R3 V 24 V P3 = R3 = 2.0 Ω + 4.0 Ω + 6.0 Ω ( 6.0 Ω ) = 24 W R1 + R2 + R3 ______________________________________________________________________________ 1128 ELECTRIC CIRCUITS 117. SSM REASONING The power P delivered to the circuit is, according to Equation 20.6c, 2 P = V / R12345 , where V is the voltage of the battery and R12345 is the equivalent resistance of the five-resistor circuit. The voltage and power are known, so that the equivalent resistance can be calculated. We will use our knowledge of resistors wired in series and parallel to evaluate R12345 in terms of the resistance R of each resistor. In this manner we will find the value for R. SOLUTION First we note that all the resistors are equal, so R1 = R2 = R3 = R4 = R5 = R. We can find the equivalent resistance R12345 as follows. The resistors R3 and R4 are in ser...
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