Using KVL in loop ABEFA
3 - i2 R2 + i1 R1 = 12
i1 R1 - i2 R2 = 9
4.90 i1 - 3.80 i2 = 9
i2 = (4.90 i1 - 9)/3.8 Eq-1
Using KVL in loop BCDEB
18 + (i1 + i2) R4 + (i1 + i2) R3 + i2 R2 = 3
18 + (i1 + i2) (R4 + R3) + i2 R2 = 3
18 + (i1 + i2) (2.20 + 3.40) + 3.80 i2 = 3
(5.6) (i1 + i2) + 3.80 i2 = - 15
5.6 i1 + 9.4 i2 = - 15
Using Eq-1
5.6 i1 + 9.4 ((4.90 i1 - 9)/3.8) = - 15
i1 = 0.41 A
Using eq-1
i2 = (4.90 (0.41) - 9)/3.8
i2 = - 1.84 A (negative sign indicates that the direction of current is opposite ti what we have assumed)
i1 + i2 = 0.41 - 1.84 = - 1.43 A (negative sign indicates that the direction of current is opposite ti what we have assumed)
= i1 R1 = (0.41) (4.90) = 2.009 volts
= i2 R2 = (1.84) (3.80) = 6.992 volts
= (i1 + i2) R3 = (1.43) (3.40) =
4.862 volts
= (i1 + i2) R4 = (1.43) (2.20) =
3.146 volts
10. -76.66 points SERCP10 18.P.029.ML.FB. Find the potential difference across each resistor in the figure below....
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a.) Find the current through resistor a) in the figure.b.) Find the potential difference across resistor a) in the figure.c.) Find the current through resistor b) in the figure.d.) Find the potential difference across resistor b) in the figure.e.) Find the current through resistor c) in the figure.f.) Find the potential difference across resistor c) in the figure.g.) Find the current through resistor d) in the figure.h.) Find the potential difference across resistor d) in the figure.I can tell that...
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