CHAPTER 26 HOMEWORK SOLUTIONS

Transcription

1 CHAPTE 6 HOMEWOK SOLUTIONS 6.. IDENTIFY: 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 uivalent parallel resistance between ab. SET UP: We use the formula for resistors in parallel. EXECUTE: 1/(.00 Ω) = 1/X + 1/(15.0 Ω) + 1/(5.0 Ω) + 1/(10.0 Ω), so X = 7.5 Ω. EVALUATE: X is greater than the uivalent parallel resistance of.00 Ω. 6.8.IDENTIFY: Eq.(6.) gives the uivalent resistance of the three resistors in parallel. For resistors in parallel, the voltages are the same and the currents add. (a) SET UP: The circuit is sketched in Figure 6.8a. EXECUTE: parallel Figure 6.8a (b) For resistors in parallel the voltage is the same across each and ual to the applied voltage; V1 V V3 E 8.0 V V1 8.0 V V I so I A V 8.0 V V3 8.0 V I 11.7 A and I3 5.8 A (c) The currents through the resistors add to give the current through the battery: I I1I I A 11.7 A 5.8 A 35.0 A EVALUATE: Alternatively, we can use the uivalent resistance Figure 6.8b E I V I E 35.0 A, which checks as shown in Figure 6.8b. (d) As shown in part (b), the voltage across each resistor is 8.0 V. (e) IDENTIFY and SET UP: We can use any of the three expressions for P: PVI I V /. They will all give the same results, if we keep enough significant figures in intermediate calculations. EXECUTE: Using P V /, 8.0 V 8.0 V P V / 490 W, / 37 W, and 1.60 P V V P3 V3 / W 4.80 EVALUATE: The total power dissipated is Pout P1 P P3 980 W. This is the same as the power Pin E I.80 V 35.0 A 980 W delivered by the battery. (f) P V /. The resistors in parallel each have the same voltage, so the power P is largest for the one with the least resistance.

2 6.9. IDENTIFY: For a series network, the current is the same in each resistor and the sum of voltages for each resistor uals the battery voltage. The uivalent resistance is 1 3. P I. SET UP: Let ,.40, EXECUTE: (a) V 8.0 V (b) I 3.18 A 8.80 (c) I 3.18 A, the same as for each resistor. (d) V1 I1 (3.18 A)(1.60 ) 5.09 V. V I (3.18 A)(.40 ) 7.63 V. V3 I3 (3.18 A)(4.80 ) 15.3 V. Note that V1V V3 8.0 V. (e) P1 I 1 (3.18 A) (1.60 ) 16. W. P I (3.18 A) (.40 ) 4.3 W. P3 I 3 (3.18 A) (4.80 ) 48.5 W. (f) Since P I and the current is the same for each resistor, the resistor with the greatest dissipates the greatest power. EVALUATE: When resistors are connected in parallel, the resistor with the smallest dissipates the greatest power IDENTIFY: Apply Ohm's law to each resistor. SET UP: For resistors in parallel the voltages are the same and the currents add. For resistors in series the currents are the same and the voltages add. EXECUTE: From Ohm s law, the voltage drop across the 6.00 resistor is V = I = (4.00 A)(6.00 ) = 4.0 V. The voltage drop across the 8.00 resistor is the same, since these two resistors are wired in parallel. The current through the 8.00 resistor is then I = V/ = 4.0 V/8.00 = 3.00 A. The current through the 5.0 resistor is the sum of these two currents: 7.00 A. The voltage drop across the 5.0 resistor is V = I = (7.00 A)(5.0 ) = 175 V, and total voltage drop across the top branch of the circuit is 175 V V = 199 V, which is also the voltage drop across the 0.0 resistor. The current through the 0.0 resistor is then I V/ 199 V/ A. EVALUATE: The total current through the battery is 7.00 A 9.95 A A. Note that we did not need to calculate the emf of the battery IDENTIFY: Apply the junction rule at points a, b, c and d to calculate the unknown currents. Then apply the loop rule to three loops to calculate E1, E and. (a) SET UP: The circuit is sketched in Figure 6.3. Figure 6.3 EXECUTE: Apply the junction rule to point a: 3.00 A 5.00 A I3 0 I A

3 Apply the junction rule to point b:.00 A I A 0 I A Apply the junction rule to point c: I3 I4 I5 0 I5 I3 I A 1.00 A 7.00 A EVALUATE: As a check, apply the junction rule to point d: I5.00 A 5.00 A 0 I A (b) EXECUTE: Apply the loop rule to loop (1): E 1 I3 E V 8.00 A V Apply the loop rule to loop (): E 5.00 A6.00 I E 30.0 V 8.00 A V.00 A E E 0 (c) Apply the loop rule to loop (3): A E E V 36.0 V A.00 A EVALUATE: Apply the loop rule to loop (4) as a check of our calculations:.00 A 3.00 A A A V 30.0 V V 18.0 V IDENTIFY: The capacitors, which are in parallel, will discharge exponentially through the resistors. SET UP: Since V is proportional to Q, V must obey the same exponential uation as Q, V = V 0 e t/c. The current is I = (V 0 /) e t/c. EXECUTE: (a) Solve for time when the potential across each capacitor is 10.0 V: t = C ln(v/v 0 ) = (80.0 Ω)(35.0 µf) ln(10/45) = 410 µs = 4.1 ms (b) I = (V 0 /) e t/c. Using the above values, with V 0 = 45.0 V, gives I = 0.15 A. EVALUATE: Since the current and the potential both obey the same exponential uation, they are both reduced by the same factor (0.) in 4.1 ms IDENTIFY: When the capacitor is fully charged the voltage V across the capacitor uals the battery t/ C emf and Q CV q Q 1 e. SET UP:. For a charging capacitor, ln e x EXECUTE: (a) x QCV 6 4 ( F)(8.0 V) C. t/ C t/ C q t (b) qq(1 e ), so e 1 and. After Q C ln(1 q / Q ) s t 310 s: ( F)(ln(1 110/165)) (c) If the charge is to be 99% of final value: q Q e t/ C (1 ) gives t C q Q 6 ln(1 / ) (463 ) ( F) ln(0.01) s. EVALUATE: The time constant is C.73 ms. The time in part (b) is a bit more than one time constant and the time in part (c) is about 4.6 time constants IDENTIFY: Apply the loop and junction rules. SET UP: Use the currents as defined on the circuit diagram in Figure 6.64 and obtain three uations to solve for the currents.

4 EXECUTE: Left loop: 14 I1 ( I1 I) 0 and 3I1 I 14. Top loop : ( I I1) I I1 0 and I 3I1 I 0. Bottom loop : ( I I1 I) ( I1 I) I 0 and I 3I1 4I 0. Solving these uations for the currents we find: I Ibattery 10.0A; I1 I 6.0A; I I.0A. 1 3 So the other currents are: I I I 4.0 A; I I I 4.0 A; I I I I 6.0 A. (b) 14.0 V V I 10.0 A EVALUATE: It isn t possible to simplify the resistor network using the rules for resistors in series and parallel. But the uivalent resistance is still defined by V I. Figure IDENTIFY and SET UP: Just after the switch is closed the charge on the capacitor is zero, the voltage across the capacitor is zero and the capacitor can be replaced by a wire in analyzing the circuit. After a long time the current to the capacitor is zero, so the current through 3 is zero. After a long time the capacitor can be replaced by a break in the circuit. EXECUTE: (a) Ignoring the capacitor for the moment, the uivalent resistance of the two parallel resistors is ; In the absence of the capacitor, the total current in the circuit (the current through the 8.00 resistor) would be 4.0 V i E 4.0 A , of which 3, or.80 A, would go through the 3.00 resistor and 1 3, or 1.40 A, would go through the 6.00 resistor. Since the current through the capacitor is given V tc by i e, at the instant t 0 the circuit behaves as through the capacitor were not present, so the currents through the various resistors are as calculated above. (b) Once the capacitor is fully charged, no current flows through that part of the circuit. The 8.00 and the 6.00 resistors are now in series, and the current through them is ie (4.0 V) /( ) 3.00 A. The voltage drop across both the 6.00 resistor and the capacitor is thus V i (3.00 A)(6.00 ) 18.0 V. (There is no current through the 3.00 resistor and so no voltage drop across it.) The charge on the capacitor is 6 5 QCV ( F)(18.0 V) C. EVALUATE: The uivalent resistance of and 3 in parallel is less than 3, so initially the current through 1 is larger than its value after a long time has elapsed.

5 6.74. IDENTIFY: With S open and after uilibrium has been reached, no current flows and the voltage across each capacitor is 18.0 V. When S is closed, current I flows through the 6.00 and 3.00 resistors. SET UP: With the switch closed, a and b are at the same potential and the voltage across the 6.00 resistor uals the voltage across the 6.00 F capacitor and the voltage is the same across the 3.00 F capacitor and 3.00 resistor. EXECUTE: (a) With an open switch: Vab E 18.0 V. (b) Point a is at a higher potential since it is directly connected to the positive terminal of the battery. (c) When the switch is closed 18.0 V I ( ). I.00 A and Vb (.00 A)(3.00 ) 6.00 V. 6 5 (d) Initially the capacitor s charges were Q3 CV ( F)(18.0 V) C and 6 4 Q6 CV ( F)(18.0 V) C. After the switch is closed 6 5 Q3 CV ( F)(18.0 V 1.0 V) C and Q6 CV ( F)(18.0 V 6.0 V) C. Both capacitors lose C. EVALUATE: The voltage across each capacitor decreases when the switch is closed, because there is then current through each resistor and therefore a potential drop across each resistor.