- #1
faoltaem
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The values for the resistors are: R1=7[tex]\Omega[/tex], R2=5[tex]\Omega[/tex], R3=4[tex]\Omega[/tex] (see diagram for placement of the resistors)
1) Suppose there is a current of 0.6A going through R1 and that the voltage supplied by the battery is 9V, determine the value of R4
2) Using the information above, determine the current through R3 and the voltage across R4
1) [tex]\frac{1}{R_{eq(3+4)}}[/tex] = [tex]\frac{1}{4}[/tex] + [tex]\frac{1}{R4}[/tex]
= [tex]\frac{R4}{4R4}[/tex] + [tex]\frac{4}{4R4}[/tex]
= [tex]\frac{R4 + 4}{4R4}[/tex]
therefore R[tex]_{eq(3+4)}[/tex] = [tex]\frac{4R4}{R4 + 4}[/tex]
R[tex]_{tot}[/tex] = R1 + R2 + R[tex]_{eq(3+4)}[/tex]
15 = 7 + 5 + [tex]\frac{4R4}{R4 + 4}[/tex]
[tex]\frac{4R4}{R4 + 4}[/tex] = 3
4R4 = 3(R4 + 4)
= 3R4 + 12
therefore R4=12[tex]\Omega[/tex]
2) I = [tex]\frac{\epsilon}{R}[/tex] = 9/15 = 0.6A
V = IR = 0.6 x 12 = 7.2V
Is this correct?
1) Suppose there is a current of 0.6A going through R1 and that the voltage supplied by the battery is 9V, determine the value of R4
2) Using the information above, determine the current through R3 and the voltage across R4
1) [tex]\frac{1}{R_{eq(3+4)}}[/tex] = [tex]\frac{1}{4}[/tex] + [tex]\frac{1}{R4}[/tex]
= [tex]\frac{R4}{4R4}[/tex] + [tex]\frac{4}{4R4}[/tex]
= [tex]\frac{R4 + 4}{4R4}[/tex]
therefore R[tex]_{eq(3+4)}[/tex] = [tex]\frac{4R4}{R4 + 4}[/tex]
R[tex]_{tot}[/tex] = R1 + R2 + R[tex]_{eq(3+4)}[/tex]
15 = 7 + 5 + [tex]\frac{4R4}{R4 + 4}[/tex]
[tex]\frac{4R4}{R4 + 4}[/tex] = 3
4R4 = 3(R4 + 4)
= 3R4 + 12
therefore R4=12[tex]\Omega[/tex]
2) I = [tex]\frac{\epsilon}{R}[/tex] = 9/15 = 0.6A
V = IR = 0.6 x 12 = 7.2V
Is this correct?