Physics Solution Manual for 1100 and 2101

The two resistors are wired in parallel so we can

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Unformatted text preview: istor. The directions of the currents are arbitrary, and should they be incorrect, the currents will turn out to be negative quantities. Having chosen the currents, we also mark the ends of the resistors with the plus and minus signs that indicate that the currents are directed from higher (+) toward lower (−) potential. These plus and minus signs will guide us when we apply Kirchhoff’s loop rule. Chapter 20 Problems A + 2.00 − I1 − E I2 − + + − 3.00 V + 8.00 B 1109 6.00 V + − − I3 9.00 V 4.00 + D F C SOLUTION Applying the junction rule to junction B, we find I1 + I 3 = 123 44 Into junction (1) I2 123 Out of junction Applying the loop rule to loop ABCD (going clockwise around the loop), we obtain I1 ( 2.00 Ω ) = 6.00 V + I 3 ( 4.00 Ω ) + 3.00 V 14 3 14444 244444 24 4 3 Potential drops (2) Potential rises Applying the loop rule to loop BEFC (going clockwise around the loop), we obtain I 2 ( 8.00 Ω ) + 9.00 V + I3 ( 4.00 Ω ) + 6.00 V = 0 24 14444444 244444444 14 3 4 3 Potential drops (3) Potential rises Substituting I2 from Equation (1) into Equation (3) gives ( I1 + I3 ) (8.00 Ω ) + 9.00 V + I3 ( 4.00 Ω ) + 6.00 V = 0 I1 ( 8.00 Ω ) + I3 (12.00 Ω ) + 15.00 V = 0 Solving Equation (2) for I1 gives I1 = 4.50 A + I3 ( 2.00 ) This result may be substituted into Equation (4) to show that (4) 1110 ELECTRIC CIRCUITS 4.50 A + I 3 ( 2.00 ) ( 8.00 Ω ) + I 3 (12.00 Ω ) + 15.00 V = 0 I3 ( 28.00 Ω ) + 51.00 V = 0 I3 = or −51.00 V = −1.82 A 28.00 Ω The minus sign indicates that the current in the 4.00-Ω resistor is directed downward , rather than upward as selected arbitrarily in the drawing. 85. SSM REASONING We begin by labeling the currents in the three resistors. The drawing below shows the directions chosen for these currents. The directions are arbitrary, and if any of them is incorrect, then the analysis will show that the corresponding value for the current is negative. 5.0 Ω + – – + I1 10.0 V + 15.0 V – – + I2 10.0 Ω I3 + 10.0 Ω – + 2.0 V – We then mark the resistors with the plus and minus signs that serve as an aid in identifying the potential drops and rises for the loop rule, recalling that conventional current is always directed from a higher potential (+) toward a lower potential (–). Thus, given the directions chosen for I1, I2 , and I3 , the plus and minus signs must be those shown in the drawing. We can now use Kirchhoff's rules to find the voltage across the 5.0-Ω resistor. SOLUTION Applying the loop rule to the left loop (and suppressing units for convenience) gives 5.0 I1 + 10.0 I 3 + 2.0 = 10.0 (1) Similarly, for the right loop, 10.0 I 2 + 10.0 I 3 + 2.0 = 15.0 (2) If we apply the junction rule to the upper junction, we obtain I1 + I 2 = I 3 Subtracting Equation (2) from Equation (1) gives (3) Chapter 20 Problems 5.0 I1 – 10.0 I2 = –5.0 1111 (4) We now multiply Equation (3) by 10 and add the result to Equation (2); the result is 10.0 I1 + 20.0 I2 = 13.0 (5) If we then multiply Equation (4) by 2 and add the result to Equation (5), we obtain 20.0 I1 = 3.0 , or solving for I1 , we obtain I1 = 0.15 A . The fact that I1 is positive means that the current in the drawing has the correct direction. The voltage across the 5.0-Ω resistor can be found from Ohm's law: V = (0.15 A)(5.0 Ω) = 0.75 V Current flows from the higher potential to the lower potential, and the current through the 5.0-Ω flows from left to right, so the left end of the resistor is at the higher potential. ______________________________________________________________________________ 86. REASONING a. Currents I5 and I2 both go into junction C (see the drawing). This combined current I5 + I2 passes through the battery and splits up again at junction A, with I1 = 9.0 A going to the resistor R1, and I3 = 12.0 A going to the R1 = 4.0 Ω A resistor R3. By Kirchhoff’s junction − + rule, therefore, the sum of the first pair I1 = 9.0 A + V = 75.0 V of currents must equal the sum of the second pair of currents. − + R2 b. We will apply Kirchhoff’s loop rule B to loop CAB (see the drawing) and solve + − for the resistance R2, which is the only I2 = 6.0 A unknown variable that appears in the loop rule. The + and − signs on either R4 = 2.0 Ω end of the resistors indicate that current flows from higher potential (+) to lower potential (−). They do not indicate that the ends of the resistors are charged in any way. I3 = C R3 12.0 A − − R5 = 2.2 Ω I5 + D E c. We will find R3 by writing out Kirchhoff’s loop rule for the loop CAED (see the drawing). SOLUTION a. As noted in the REASONING, the sum of the currents flowing into the battery must equal the sum of the currents flowing out of the battery: 1112 ELECTRIC CIRCUITS I 2 + I5 = I1 + I3 1 3 13 2 2 into the battery I5 = I1 + I3 − I 2 = 9.0 A + 12.0 A − 6.0 A = 15.0 A so out of the battery b. To apply Kirchhoff’s loop rule to loop CAB (see the drawing), we imagine traversing the loop counter-clockwise. Observing the + and − signs encountered along the...
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