No credit will be given for answers without appropriate supporting work. Minimum

Question

No credit will be given for answers without appropriate supporting work. Minimum good presentation requires the following: (1) Symbolic expression for any formula, (2) Manipulation of symbolic expressions, not numeric expressions, (3) Substitution of numbers with units, (4) Reporting final answers with correct units and vector expressions, (5) Enough English description to allow the reader to have some idea what you are doing without looking at the math.

Open-Response Homework Problem 8.3 A 12V battery is connected to two parallel combinations of resistors. The two parallel combinations are connected in series. The first parallel combination, the one at the top of the page, is made up of a 200?, 300?, and a 500? resistor. The second combination is made up of three 300? resistors. The two parallel combinations are connected in series with the battery. (a)Compute the equivalent resistance, Req, of the circuit. (b)Compute the current drawn by the circuit. (c)Compute the power dissipated by the 200? resistor.

Explanation / Answer

Here ,

for the resistances in parallel

1/Rp = 1/R1 + 1/R2 + 1/R3

for the resistances in series

Rs = R1 + R2

a) for the top resistances in parallel

1/Rp1 =1/200 + 1/300 + 1/500

Rp1 = 96.8 Ohm

for the lower resistances

1/Rp2 = 1/300 + 1/300 + 1/300

Rp2 = 100 Ohm

Now, for Rp1 and Rp2 in series

Req = 100 + 96.8

Req = 196.8 Ohm

the equivalent resistance of the circuit is 196.8 Ohm

b)

Let the current drawn is I

Using Ohm's law

I = V /Req

I = 12/196.8 A

I = 0.061 A

the current drawn from the battery is 0.061 A

c)

Now, for voltage across 200 Ohm

V1 = I * Rp1

V1 = 0.061 * 96.8

V1 = 5.903 V

power dissipated in 200 Ohm = V1^2/R

power dissipated in 200 Ohm = 5.903^2/(200)

power dissipated in 200 Ohm = 0.174 W

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