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Unformatted text preview: e of R234? (A) R234 = 1 Ω (B) R234 = 2 Ω (C) R234 = 4 Ω (D) R234 = 6 Ω (1/Rparallel) = (1/Ra) + (1/Rb) 1/R234 = (1/3) + (1/6) = (3/6) Ω −1 R234 = 2 Ω BB Calculation
V In the circuit shown: V = 18V, I1 = I234 R234 R1 = 1Ω, R24 = 6Ω R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. R234 = 2Ω What is V2, the voltage across R2?
R1 and R234 are in series. R1234 = 1 + 2 = 3 Ω V = I1234 R1234 Ohm’s Law tells us I1234 = V/R1234 = 18 / 3 = 6 Amps Calculation
V I1234 R1234 In the circuit shown: V = 18V, R1 = 1Ω, R24 = 6Ω R2 = 2Ω, R3 = 3Ω, and R4 = 4Ω. R234 = 2Ω What is V2, the voltage across R2?
R1 a V I234 = I1234 Since R1 in series w/ R234 R234 = I1 = I234 V234 = I234 R234
= 6 x 2 b = 12 Volts • What is Vab, the voltage across R234 ? (A) Vab = 1 V (B) Vab = 2 V (C) Vab = 9 V (D) Vab = 12 V (E) Vab = 16 V I1234 = 6 A BB Calculation
Calculation R1 V R234 R1 V R3 Which of the following is/are true?
A) V234 = V24 B) I234 = I24 C) Both A+B D) None Since R3 and R24 where combined in parallel to get R234 Voltages are same!
I24 = V24 / R24 = 12 / 6 = 2 Amps R24 V = 18V R1 = 1Ω
R2 = 2Ω
R3 = 3Ω
R4 = 4Ω. R24 = 6...
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- Spring '14