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In the figure R1 = 120 N, R2 = R3 = 66.0 N, R4 = 71.7 Q, and the ideal battery has emf ɛ = 6.00 V. (a) What is the equivalent resistance? What is i in (b) resistance 1, (c) resistance 2, (d) resistance 3, and (e) resistance 4? R ww

In the figure R1 = 120 N, R2 = R3 = 66.0 N, R4 = 71.7 Q, and the ideal battery has emf ɛ = 6.00 V. (a) What is the equivalent resistance? What is i in (b) resistance 1, (c) resistance 2, (d) resistance 3, and (e) resistance 4? R ww

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In the figure R1 = 120 Q, R2 = R3 = 66.0 Q, R4 = 71.7 Q, and the ideal battery has emf ɛ = 6.00 V. (a) What is the equivalent resistance? What is i in (b) resistance 1, (c) resistance 2, (d) resistance 3, and (e) resistance 4? RA R2 R (a) Number Units (b) Number i Units (c) Number Units (d) Number i Units (e) Number i Units > > >
Transcribed Image Text:In the figure R1 = 120 Q, R2 = R3 = 66.0 Q, R4 = 71.7 Q, and the ideal battery has emf ɛ = 6.00 V. (a) What is the equivalent resistance? What is i in (b) resistance 1, (c) resistance 2, (d) resistance 3, and (e) resistance 4? RA R2 R (a) Number Units (b) Number i Units (c) Number Units (d) Number i Units (e) Number i Units > > >
or in parallel to make a 10 N resist- ance that is capable of dissipating at 1190 CHAPTER 27 R1 least 5.0 W? R4 R3 problem demands Ptotal 2 5.0P, so n² must be at least as large as 5.0. Since n must be an integer, the smallest it can be is 3. The least number of resistors is n = 9. R2 In Fig. 27-53, RỊ 50.0 N, R4 = 75.0 N, and the ideal battery has emf E = 6.00 V. (a) What is the equivalent resistance? What is i in (b) resistance 1, (c) resist- ance 2, (d) resistance 3, and (e) resist- •44 GO 100 N, R2 = R3 44. (a) Resistors R2, R3, and R4 are in parallel. By finding a common denominator and simplifying, the equation 1/R = 1/R2 + 1/R3 + 1/R4 gives an equivalent resistance of Figure 27-53 Problems 44 and 48. R,R,R, (50.02)(50.02)(75.02) R= RR +R,R, + RR, (50.0Ω)(50.0Ω) + (50.0Ω)(75.0Ω) + (50.0Ω)(750Ω) =18.8N. ance 4? R1 R1 Thus, considering the series contribution of resistor R1, the equivalent resistance for the network is Rea = R1 + R = 100 Q + 18.8 Q =118.8 Q × 119 Q. •45 ILW In Fig. 27-54, the resistances 1.0 N and R2 = 2.0 N, are R1 and the ideal batteries have emfs R2 (b) i¡ = ɛ/Req = 6.0 V/(118.8 N) = 5.05 × 10² A. R1 E1 = 2.0 V and E, = Ez = 4.0 V. What are the (a) size and (b) direction (up or down) of the current in battery 1, the (c) size and (d) direction of the (c) iz = (ɛ- V1)/R2 = (ɛ- i¡R¡)/R2 = [6.0V – (5.05 × 10² A)(1002)]/50 2 = 1.90 × 10² A. R1 (d) iz = (ɛ- Vi)/R3 = i,R2/R3 = (1.90 × 10² A)(50.0 2/50.0 N) = 1.90 × 10² A. (e) i4 = i1 – i2 – iz = 5.05 × 10² A – 2(1.90 × 10² A) = 1.25 x 10² A. Figure 27-54 Problem 45.
Transcribed Image Text:or in parallel to make a 10 N resist- ance that is capable of dissipating at 1190 CHAPTER 27 R1 least 5.0 W? R4 R3 problem demands Ptotal 2 5.0P, so n² must be at least as large as 5.0. Since n must be an integer, the smallest it can be is 3. The least number of resistors is n = 9. R2 In Fig. 27-53, RỊ 50.0 N, R4 = 75.0 N, and the ideal battery has emf E = 6.00 V. (a) What is the equivalent resistance? What is i in (b) resistance 1, (c) resist- ance 2, (d) resistance 3, and (e) resist- •44 GO 100 N, R2 = R3 44. (a) Resistors R2, R3, and R4 are in parallel. By finding a common denominator and simplifying, the equation 1/R = 1/R2 + 1/R3 + 1/R4 gives an equivalent resistance of Figure 27-53 Problems 44 and 48. R,R,R, (50.02)(50.02)(75.02) R= RR +R,R, + RR, (50.0Ω)(50.0Ω) + (50.0Ω)(75.0Ω) + (50.0Ω)(750Ω) =18.8N. ance 4? R1 R1 Thus, considering the series contribution of resistor R1, the equivalent resistance for the network is Rea = R1 + R = 100 Q + 18.8 Q =118.8 Q × 119 Q. •45 ILW In Fig. 27-54, the resistances 1.0 N and R2 = 2.0 N, are R1 and the ideal batteries have emfs R2 (b) i¡ = ɛ/Req = 6.0 V/(118.8 N) = 5.05 × 10² A. R1 E1 = 2.0 V and E, = Ez = 4.0 V. What are the (a) size and (b) direction (up or down) of the current in battery 1, the (c) size and (d) direction of the (c) iz = (ɛ- V1)/R2 = (ɛ- i¡R¡)/R2 = [6.0V – (5.05 × 10² A)(1002)]/50 2 = 1.90 × 10² A. R1 (d) iz = (ɛ- Vi)/R3 = i,R2/R3 = (1.90 × 10² A)(50.0 2/50.0 N) = 1.90 × 10² A. (e) i4 = i1 – i2 – iz = 5.05 × 10² A – 2(1.90 × 10² A) = 1.25 x 10² A. Figure 27-54 Problem 45.
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