The circuit shown here can\'t be reduced, since none of the resistors are connec

Question

The circuit shown here can't be reduced, since none of the resistors are connected in series or in parallel with each other. The unlabeled arrows show the direction of current in resistors 2, 4, and 5. Use Kirchhoffs junction or loop rules to answer questions (a)-(c) below. Show work for all calculations, and briefty explain your reasoning on current direction in (a) and (b). The symbols h, h, etc. represent the currents in resistors Ri, Rs, etc. (a) Ifh=1.60 A and 1,-2.30 A, what is 11, and does it flow from b to c or from c to b? (b) If I-1.52 A, what is I, and does it flow from a to b or from b to a? (c) What is the battery's emfAV (d) What is the battery's current h? (e) Based on your answers to (c) and (d) what is the circuit's equivalent resistance? R1 = 8.0 R= 2.0 ] R, 12 Rs 6.0

Explanation / Answer

A) using KCL, current flows from b to c.

i3+i2 =i5

i3 = 2.3-1.6 = 0.7 A

B) Again using KCL,

i3+i4 = i1

i1 = 0.7+1.52 = 2.22A

C) V = i5R5 + i2R2 = 2.3*6 +1.6*12

= 33 V

D) ib = i4+i5 = 1.52+2.30 = 3.82 A

E) equivalent resistance = V/ib = 33/3.82

= 8.64 ohm

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