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

Question

2. 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. 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

Explanation / Answer

(a) The resistances 60 ohm and 40 ohm are in parallel.

So, its equivalent = (60*40) / (60+40) = 24 ohm

again -

resistances 60 ohm, 60 ohm and 45 ohm are in parallel.

So, its equivalent -

equivalent of 60||60 = 30 ohm

and 30||45 = (30*45) / (30+45) = 18 ohm

Now these two combinations are in series.

So, the equivalent = 24 + 18 = 42 ohm.

Now this resistance is in parallel with 21 ohm.

So, the total resistance between a and b = (42*21) / (42+21) = 14 ohm.

(b) total current = 20 / 14 = 1.43 A

Current through the left branch = 1.43 - 1.43*42 / (42+21) = 1.43 - 0.95 = 0.48 A

Potential difference across the left upper combination = 0.48*24 = 11.52 V

So, potential difference across the 45 ohm resistor = 20 - 11.52 = 8.48 V

Current = 8.48/45 = 0.19 A

(c) Maximum resistance will occure when the resistances are connected in series.

its value = 60 + 40 + 60 + 60 + 45 + 21 = 286 ohm

(d) Minimum value of resistance will occure when connected in parallel.

Lets its value be R.

So -

1/R = 1/60 + 1/40 + 1/60 + 1/60 + 1/45 + 1/21 = 0.017 + 0.025 + 0.017 + 0.017 + 0.022 + 0.048

=> 1/R = 0.146

=> R = 6.85 ohm.

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