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2-47 Use current division in the circuit of Figure P2-47 to find Rx so that the voltage out is 3 V. Repeat for 5 V. ΙΑ O Rx www 102 102 +21 VO

2-47 Use current division in the circuit of Figure P2-47 to find Rx so that the voltage out is 3 V. Repeat for 5 V. ΙΑ O Rx www 102 102 +21 VO

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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2.47 2-47 Use current division in the circuit of Figure P2-47 to find Rx so that the voltage out is 3 V. Repeat for 5 V. Rx ΤΑ 1092 FIGURE P2-47 + 1002 VO EXAMPLE 2-5 Given i = +4 A, i=+1 A, 4+2 A in the circuit shown in Figure 2-11, find i2 and is. SOLUTION: Using the node A constraint in Eq. (2-8) yields -1₁-2=-(+4)-iz=0 The sign outside the parentheses comes from the node A KCL connection constraint in Eq. (2-8). The sign inside the parentheses comes from the actual direction of the current. Solving this equation for the unknown current, we find that i = -4 A. In this case, the minus sign indicates that the actual direction of the current 2 is directed upward in Figure 2-11, which is opposite to the reference direction assigned. Using the second KCL equation in Eq. (2-8), we can write in-is-i +i5-(+4)-(+1)-(+2) -=0 which yields the result is = -1 A. Again, the signs inside the parentheses are associated with the actual direction of the current, and the signs outside come from the node B KCL connection constraint in Eq. (2-8). The minus sign in the final answer means that the current is is directed in the opposite direction from its assigned reference direction. We can check our work by substituting the values found into the node C constraint in Eq. (2-8). These sub- stitutions yield +12+13+14-15=(-4)-(-1)+(+2)-(-1)=0 as required by KCL. Given three currents, we determined all the remaining currents in the circuit using only KCL without knowing the element constraints.
Transcribed Image Text:2.47 2-47 Use current division in the circuit of Figure P2-47 to find Rx so that the voltage out is 3 V. Repeat for 5 V. Rx ΤΑ 1092 FIGURE P2-47 + 1002 VO EXAMPLE 2-5 Given i = +4 A, i=+1 A, 4+2 A in the circuit shown in Figure 2-11, find i2 and is. SOLUTION: Using the node A constraint in Eq. (2-8) yields -1₁-2=-(+4)-iz=0 The sign outside the parentheses comes from the node A KCL connection constraint in Eq. (2-8). The sign inside the parentheses comes from the actual direction of the current. Solving this equation for the unknown current, we find that i = -4 A. In this case, the minus sign indicates that the actual direction of the current 2 is directed upward in Figure 2-11, which is opposite to the reference direction assigned. Using the second KCL equation in Eq. (2-8), we can write in-is-i +i5-(+4)-(+1)-(+2) -=0 which yields the result is = -1 A. Again, the signs inside the parentheses are associated with the actual direction of the current, and the signs outside come from the node B KCL connection constraint in Eq. (2-8). The minus sign in the final answer means that the current is is directed in the opposite direction from its assigned reference direction. We can check our work by substituting the values found into the node C constraint in Eq. (2-8). These sub- stitutions yield +12+13+14-15=(-4)-(-1)+(+2)-(-1)=0 as required by KCL. Given three currents, we determined all the remaining currents in the circuit using only KCL without knowing the element constraints.
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