Determine the magnitudes of the currents in each resistor . The batteries have e
ID: 2280234 • Letter: D
Question
Determine the magnitudes of the currents in each resistor . The batteries have emfs of E1=9.0V and E2=12.0Vand the resistors have values of R1=25?, R2=18?, and R3=35?.
I1, I2, I3?
R1
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l I
I I
E1= R2 I
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I I
I >
E2= R3<
I >
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Explanation / Answer
I would set this problem up as two loop equations of the possible three. Let us take the first loop as being the current flow clockwise through E1, R1, and R2. The second loop we take as counterclockwise through E2, R3, and R2. The third loop would be through E1, E2, R3 and R1, but we don't need this one because it is just the second loop minus the first.
Let us call the current through the first loop I1 because this is the only current flowing through R1, with positive I1 being rightward through R1, and the current through the second loop I3 because this is the only current flowing through R3, with positive I3 being upward through R3. If we take I2 as being positive in the direction to the left, then I2 = I3 + I1 Taking the voltage rises and drops around the first loop, we get
E1 ? I1R1 ? (I1+I3)R2 = 0
and a trip around the second loop gives us
E2 ? I3R3 ? (I1+I3)R2 = 0
We solve the simultaneous linear equations for I1 and I3 to get
I1 = [E1(R2+R3) ? E2R2] / (R1R2 + R1R3 + R2R3)
I3 = [E2(R1+R2) ? E1R2] / (R1R2 + R1R3 + R2R3)
and
I2 = I1 + I3
= (E1R3 + E2R1) / (R1R2 + R1R3 + R2R3)
In this problem, the denominators will be
D = 15*20 + 15*31 + 20*31 = 1385
and the currents will be
I1 = [9*(20+31)?15*20] / D
= 159/1385 = 0.115 A (rightward)
I2 = (9*31 + 15*15) / D
= 504/1385
= 0.364 A (leftward)
I3 = [15*(15+20)?9*20] / D
= 345 / 1385
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