Consider the circuit shown in the figure(Figure 1). Suppose the four resistors i
ID: 1880748 • Letter: C
Question
Consider the circuit shown in the figure(Figure 1). Suppose the four resistors in this circuit have the values R1-13 . 2-800 , R3-61 , and R4-15 , and that the emf of the battery is E: 18 V Part A Find the current through each resistor using the rules for series and parallel resistors Express your answers using two significant figures separated by commas. 1,12,13, Submit uest Ans Part B gure 1 of 1 Find the current through each resistor using Kirchhoffs rules Express your answers using two significant figures separated by commas. I1, I2, I3, I4 - × Incorrect; Try Again; 6 attempts remainingExplanation / 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:
R2 and R4 are in series, So
R24 = 8 + 15 = 23 Ohm
Now R24 and R3 are in series, So
R234 = R24*R3/(R24 + R3) = 23*6.1/(23 + 6.1) = 4.8 ohm
Now R234 and R1 are in series, So
Req = R1 + R234 = 13 + 4.8 = 17.8
From Ohm's law:
ieq = V/Req = 18/17.8 = 1.011 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. So,
ieq = i1 = i234
i1 = 1.011 Amp
i234 = 1.011 Amp = 1.0 Amp
V234 = i234*R234 = 1.0*4.8 = 4.8 V
Now Since R3 and R24 are in parallel, So
V3 = V24 = V234 = 4.8 V
i3 = V3/R3 = 4.8/6.1 = 0.79 Amp
V24 = 4.8 V
i24 = i234 - i3 = 1.0 - 0.79 = 0.21 Amp
Since R2 and R4 are in series, So
i2 = i4 = i24
i2 = 0.21 Amp
i4 = 0.21 Amp
So,
i1, i2, i3, i4 = (1.0, 0.21, 0.79, 0.21) Amp
Part B.
Using Kirchoff's law:
Current through R1, towards left = i1
Current through R2 & R4, downward = i2
Current through R3, downward = i3
Using Junction law at B
i1 = i2 + i3
Now Kirchoff's Voltage law in left loop
E = i3*R3 + i1*R1
18 = 6.1*i3 + 13*i1
Now Kirchoff's Voltage law in right loop
0 = R3*i3 - R2*i2 - R4*i2
R3*i3 = (R2 + R4)*i2
6.1*i3 = (8 + 15)*i2
i3 = (23/6.1)*i2 = 3.77*i2
Using these three equation
i1 = i2 + i3
i1 = i2 + 3.77*i2
i1 = 4.77*i2
18 = 6.1*3.77*i2 + 13*4.77*i2
i2 = 18/(6.1*3.77 + 13*4.77)
i2 = 0.21 Amp
Since R2 are R4 are in series, So
i2 = i4 = 0.21 Amp
i3 = 3.77*i2 = 3.77*0.21 = 0.79 Amp
i1 = 4.77*i2 = 4.77*0.21 = 1.00 Amp
So,
i1, i2, i3, i4 = (1.0, 0.21, 0.79, 0.21) Amp
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