Chpater 17-1-3563-1r1. x / wh U2-Qua . ???? 25 \', e lf The Potential Drop Be. k

Question

Chpater 17-1-3563-1r1. x / wh U2-Qua . ???? 25 ', e lf The Potential Drop Be. kCOwww.webasi 3. 4.2 points Tipler 25 P 075 My Notes Ask Your Consider the resistor network shown in the figure below, where R-4 ? and R2-9 ? 12? (a) Hind the equivalent resistance between points a and b in Higure 26-51 (b) If the potential drop between a and h is 12 V, find the crrent in each resistor. 16 ? (upper branch) ? (lower branch) eBbook Submit Answer Save ProgressPractice Another Version -3.3 polnts Tiplkrs 25 P00 My Notes Ask Your For the circuit shown in the figure below, find the following 1.00 ? 400V 2.000 choater 1-1lec... ptxchpater 17 lecture.pptr A Show all X 12 PM 3/26/2018

Explanation / Answer

Remember:
For series combination
Req = R1 + R2 + R3 +...............
for parallel combination
1/Req = 1/R1 + 1/R2 + 1/R3 + ............
for 2 resistors in parallel it will be
Req = R1*R2/(R1+R2)
Using this Information:

6 ohm and R1 are in parallel, So

Rp1 = 6*4/(6 + 4) = 2.4 ohm

Rp1 and R2 are in series, So

Rs1 = 9 + 2.4 = 11.4 ohm

In upper branch 12 and 6 ohm are in series, So

Rs2 = 12 + 6 = 18 ohm

Now Rs1 and Rs2 are in parallel, So

Req = 18*11.4/(18 + 11.4) = 6.98 Ohm

Using Ohm'a law:

V = i*R

i = V/R = 12/6.98 = 1.72 Amp

Now remember in resistors parallel combination voltage distribution in each part will be same and in series combination current distribution in each resistor will be same.

Voltage in Rs1 and Rs2 will be equal to total voltage, Since they are in parallel,

Vs1 = Vs2 = 12 V

is2 = Vs2/Rs2

is2 = 12/18 = 0.67 Amp

Since in upper branch 12 and 6 ohm are in series, So current in them will be equal to current in Rs2

Current in 12 ohm = 0.67 Amp

Current in 6 ohm (upper branch) = 0.67 Amp

Now in lower branch

is1 = Vs1/Rs1 = 12/11.4 = 1.05 Amp

Again R2 is in series of Rs1, So current will be same

i2 = current in 9 ohm = 1.05 Amp

Voltage in R2 = i2*R2 = 1.05*9 = 9.45 V

Voltage in Rp1 = 12 - 9.45 = 2.55 V

Since R1 and 6 ohm are in parallel, So voltage in them will be equal to Rp1

V1 = 2.55 V

i1 = current in 4 ohm = V1/R1 = 2.55/4 = 0.63 Amp

Voltage in 6 ohm (lower) = 2.55 V

current in 6 ohm (lower) = V/R = 2.55/6 = 0.42 Amp

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