Consider the circuit shown in the figure below. (Assume R_1 = 12.5 Ohm and R_2 =

Question

Consider the circuit shown in the figure below. (Assume R_1 = 12.5 Ohm and R_2 = 8.00 Ohm) (a) Find the potential difference between points a and b. (b) Find the current in the 20.0-Ohm resistor. Consider the following figure. (Assume V = 13.C V, R_1 = 2.00 Ohm, and R_2 = 6.36 Ohm) a) Is it possible to reduce the circuit shown in the figure above to a single equivalent resistor connected across the battery? Explain. This answer has not been graded yet. (b) Find the current in the 2.00 - Ohm resistor. (c) Calculate the power delivered by the battery to the circuit.

Explanation / Answer

1) 8 and 20 are in series = 20+8 = 28 ohm

28 and 8 are in parellel = 28*8/36 = 6.2 ohm

6.2 and 12.5 are in parallel = 6.2*12.5/18.7 = 4.14 ohm

12.5 and 4.14 are in series = 12.5+4.14 = 16.64 ohm

i = v/R = 25/16.64 = 1.5 amp

in series i same

i across 4.14 ohm is = 1.5 amp

voltage = 1.5*4.14 = 6.21 V

in parallel v is same

vab = 6.21 V

b) v across 6.2 ohm = 6.21 V

v acorss 28 and 8 ohm = 6.21 V

I ACROSS 28 OHM = 6.21/28 = 0.221 amp

in series i same

i 20 = 0.221 amp

2) a) yes

b) 3 and 6.36 are in parallel = 2.04 ohm

2,4 and 2.04 are in series = 2+4+2.04 = 8.04 ohm

i = v/R = 13/8.04 = 1.62 amp

in series i same

so i across 2 ohm resistor is = 1.62 amp

p = v*i = 13*1.62 = 21.06 W

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