Question
Asked Dec 19, 2019
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Two resistors, R1 and R2, are connected in series. (a) If R1 =
2.00 Ω and R2 = 4.00 Ω, calculate the single resistance equivalent
to the series combination. (b) Repeat the calculation for
a parallel combination of R1 and R2.

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Expert Answer

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Given Info: The two resistors R = 2.00 N and R2 = 4.00 2 are connected in series. Explanation: Formula to calculate the single resistance equivalent to the series combination is, Requ = R1 + R2 Requ is the equivalent resistance, R and R2 is the resistors connected in series, N for R2 to find Requ- Substitute 2.00 N for R1 and 4.00 Requ = 2.00 2 +4.00 N = 6.00 N

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Conclusion: When two resistors R1 and R2 are connected in series, a single equivalent resistance is equivalent to the sum of the two individual resistors. Therefore, the single resistance equivalent for series combination is 6.00 2.

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Given Info: The two resistors R = 2.00 N and R2 = 4.00 2 are connected in parallel combination. Explanation: Formula to calculate the single resistance equivalent to the parallel combination is, R2 %3D Reg R, Requ is the equivalent resistance, • Rị and R, is the resistors connected in parallel, N for R1 and 4.00 N for R2 to find Requ- Substitute 2.00 Re 4.00 2 2.00 2 4.00 2 4.00 2 Requ = 1.33 2 %3D

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Current Electricity

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