Question

Consider the circuit shown in the figure(Figure 1). Suppose the four resistors in this circuit have the values R1 = 12 12, R2 = 7.0 N, R3 = 7.9 2 , and R4 = 12 12, and that the emf of the battery is € = 18 V You may want to review (Pages 748 - 751)

Answer 1

Paxton R2 Given, Ri=120- КА A R₂=7 R R₂ = 7.92 R = 122 C { = 18 V The equivalent resistance of the circuit Req = (R₂ + R4) 11 R3 + R, (7 + 12 11 79 + 12 19 11 7.9 + 12 11 19.8 7.9 19 + 7.9 + 12 11 5.58 + 12 17:58 2 The total urrent flowing through the circuit, E Req 1958 - 1.024 A The ensrent flowing through resistance R, is, Hy = 1.024 A

past-B w 24 C Applying KvL i, R, T for the loop , we get (11.12) R3 = 4 > (2, +23) 1 - 12 R₂ = { 19.91 - 7.92 = 18 0 Applying KvL for the look for the roof -2, we get (R2 + Ra) iz + (2-1) R₂ = 0 R - R3 il (R2 + R3 +R4) iz = - 7-9 + 26.9 2.0 + D equation 10 x +9 = 154.211 - 62.41 12 = 142.2 equation-10-199 159.213, + 535.311 = (****) 472.9 12 = 142.2 142.2 472.9 - 0.300 A from equation (1), 26.9xo.3 7.9 = 1.024DA

The current flow the resistance - wing throug Ry is is , I, = 1.024 A is , az D.300A R2 R3 is, I3 (1,-1) = (1.024-0.300) = 0.724 Ra is, Iq = 0.300 A
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