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Physics Solution Manual for 1100 and 2101

Take i3 to the right i2 to the left and i1 to the

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Unformatted text preview: way, we see that the potential rises when we cross the battery, and drops when we cross each resistor. The sum of the potential rises must equal the sum of the potential drops, so we have that V = I1R1 + I R2 { 14 2 3 24 Potential rises (1) Potential drops Solving Equation (1) for R2, we obtain I 2 R2 = V − I1 R1 or R2 = V − I1 R1 75.0 V − ( 9.0 A ) ( 4.0 Ω ) = = 6.5 Ω I2 6 .0 A c. This time, we apply the loop rule to loop CAED (see the drawing), traversing it clockwise. This procedure yields V = I 3 R3 + I 5 R5 { 14243 Potential rises (2) Potential drops Solving Equation (2) for R3, we obtain I 3 R3 = V − I 5 R5 87. or R3 = V − I 5 R5 I3 = 75.0 V − (15.0 A ) ( 2.2 Ω ) = 3.5 Ω 12.0 A R c = 50.0 Ω SSM REASONING Since only 0.100 mA out of G the available 60.0 mA is needed to cause a full-scale 59.9 mA deflection of the galvanometer, the shunt resistor must 0.100 mA B A allow the excess current of 59.9 mA to detour around Shunt the meter coil, as the drawing at the right indicates. resistor R The value for the shunt resistance can be obtained by 60.0 mA recognizing that the 50.0-Ω coil resistance and the shunt resistance are in parallel, both being connected between points A and B in the drawing. Thus, the voltage across each resistance is the same. Chapter 20 Problems 1113 SOLUTION Expressing voltage as the product of current and resistance, we find that 0.100 × 10 A ) ( 50.0 Ω ) (59.9 × 10 A ) ( R3 = (14444244443) 1444 2444 –3 –3 Voltage across shunt resistance Voltage across coil resistance ( 0.100 × 10 A ) (50.0 Ω ) = R= –3 0.0835 Ω –3 59.9 × 10 A ______________________________________________________________________________ 88. REASONING The following drawing shows a galvanometer that is being used as a nondigital voltmeter to measure the potential difference between the points A and B. The coil resistance is RC and the series resistance is R. The voltage VAB between the points A and B is equal to the voltage across the coil resistance RC plus the voltage across the series resistance R. R RC G I A B VAB SOLUTION Ohm’s law (Equation 20.2) gives the voltage across the coil resistance as I RC and that across the series resistor as I R. Thus, the voltage between A and B is VAB = I RC + I R Solving this equation for R yields VAB − I RC 10.0 V − ( 0.400 × 10−3 A )( 60.0 Ω ) = = 2.49 ×104 Ω I 0.400 ×10−3 A ______________________________________________________________________________ R= 89. REASONING According to Ohm’s law and under full-scale conditions, the voltage V across the equivalent resistance Req of the voltmeter is V = IReq, where I is the full-scale current of the galvanometer. The full-scale current is not given in the problem statement. However, it is the same for both voltmeters. By applying Ohm’s law to each voltmeter, we will be able to eliminate the current algebraically and calculate the full-scale voltage of voltmeter B. SOLUTION Applying Ohm’s law to each voltmeter under full-scale conditions, we obtain 1114 ELECTRIC CIRCUITS VA = IReq, A VB = IReq, B and Dividing these two equations and eliminating the full-scale current I give VB VA = IReq, B IReq, A = Req, B Req, A or VB = VA Req, B Req, A = ( 50.0 V ) 1.44 × 105 Ω 2.40 × 105 Ω = 30.0 V 90. REASONING AND SOLUTION The voltages across the galvanometer and shunt resistances are equal since they are in parallel, so IgRg = IsRs Ig Rs = I s or Rg where Ig = 0.150 × 10–3 A and Is = 4.00 × 10–3 A – 0.150 × 10–3 A = 3.85 × 10–3 A. Then 0.150 × 10−3 A Rs = (12.0 Ω ) = 0.468 Ω 3.85 ×10−3 A The equivalent parallel resistance can be obtained as follows: 1 1 1 = + Rp 0.468 Ω 12.0 Ω or Rp = 0.450 Ω ______________________________________________________________________________ 91. SSM REASONING AND SOLUTION For the 20.0 V scale V1 = I(R1 + Rc) For the 30.0 V scale V2 = I(R2 + Rc) Subtracting and rearranging yields I= V2 − V1 30.0 V − 20.0 V = = 8.00 × 10 –3 A R2 − R1 2930 Ω − 1680 Ω Substituting this value into either of the equations for V1 or V2 gives Rc = 820 Ω . ______________________________________________________________________________ Chapter 20 Problems 1115 92. REASONING AND SOLUTION a. According to Ohm's law (Equation 20.2, V = IR ) the current in the circuit is I= V V = R+ R 2R The voltage across either resistor is IR, so that we find V 60.0 V V IR = = 30.0 V R = = 2 2 2R b. The voltmeter’s resistance is R v = V / I = (60.0 V)/(5.00 × 10 –3 A) = 12.0 × 10 3 Ω , and this resistance is in parallel with the resistance R = 1550 Ω. The equivalent resistance of this parallel combination can be obtained as follows: 1 11 = + Rp Rv R Rp = or RRv R + Rv = (1550 Ω ) (12.0 × 103 Ω ) 1550 Ω + 12.0 × 103 Ω = 1370 Ω The voltage registered by the voltmeter is IRp, where I is the current supplied by the battery to the series combination of the other 1550-Ω resistor and Rp. According to Ohm’s law I= 60.0 V = 0.0205 A 1550 Ω + 1370 Ω Thus, the voltage registered by the voltmeter is IR p = (0.0205 A )(1370 Ω ) = 28.1 V ______________________________________________...
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