A complex circuit is shown in Figure 2. a. b. 2. What is the equivalent resistan

Question

A complex circuit is shown in Figure 2. a. b. 2. What is the equivalent resistance between points a and b? Assume a 20 V battery is connected between points a and b. Solve for the current and voltage drop across the 45 2 resistor. .Suppose you wanted to create the most resistance between points a and b, using the supplied resistors and any wire configuration you wanted. How would you assemble the resistors? Show that this resistance is greater than the resistance found in part a. d. Suppose you wanted to create the least resistance between points a and b, using the supplied resistors and any wire configuration you wanted. How would you assemble the resistors? Show that this resistance is less than the resistance found in part a. 10 S2 21 0 60f on 45 Figure 1 Figure 2

Explanation / Answer

a)
let 1/R1 = 1/60 + 1/40

R1 = 24 ohms

1/R2 = 1/60 + 1/60 + 1/45

R2 = 18 ohms

R12 = R1 + R2

= 24 + 18

= 42 ohms

1/Rab = 1/R12 + 1/R3

1/Rab = 1/42 + 1/21

Rab = 42*21/(42+21)

= 14 ohms <<<<<<<<---------------Answer

b) current through the left brach, I12 = V/R12

= 20/42

= 0.476 A

voltage drop across 45 ohm = I*R2

= 0.476*18

= 8.57 V <<<<<<<<---------------Answer

current through 45 ohms = 8.57/45

= 0.190 A <<<<<<<<---------------Answer

c) when we connect all the resistors in series the net resistance becomes maximum.

Rmax = 60 + 60 + 45 + 60 + 40 + 21

= 286 ohms <<<<<<<<---------------Answer

d) when we connect all the resistors in parallel we get least resistance.

1/R_least = 1/60 + 1/60 + 1/45 + 1/60 + 1/40 + 1/21

Rleast = 6.90 ohms <<<<<<<<---------------Answer

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