A complex circuit is shown in Figure 2. a. What is the equivalent resistance bet

Question

A complex circuit is shown in Figure 2. a. What is the equivalent resistance between points a and b? 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. C. 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 resistors and any wire configuration you wanted. How would you assemble the resistors? Show d. Suppose you wanted to create the least resistance between points a and b, using the supplied that this resistance is less than the resistance found in part a.

Explanation / Answer

Two resistors 60 and 40

R1 = 60 x 40/(60 + 40) = 24

60,60 and 45 are in parallel

1/R2 = 1/60 + 1/60 + 1/45 = 18 Ohm

R12 = 24 + 18 = 42 Ohm

Req = 42 x 21/(42 + 21) = 14 Ohm

Hence, Req = 14 Ohm

b)I = 20/14 = 1.43 A

I21 = 20/21 = 0.95 A

Ip = 1.43 - 0.95 = 0.48 A

V2 = 20 - 0.48 x 24 = 8.48 V

I45 = 8.48/45 = 0.188 A

Hence, I45 = 0.188 A

c) We can connect the resistors in series to maximise the overall resistance

R = 60 + 40 + 60 + 60 + 45 = 265

Req = 265 x 21/(265 + 21) = 19.46 Ohm

Req = 19.46 Ohm

d)All the parallel in left side

1/R1 = 1/60 + 1/40 + 1/60 + 1/60 + 1/45 = 20.29

Req = 21 x 20.29/(21 + 20.29) = 10.32

Hence, Req = 10.32 Ohm

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