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What is the resistance of a 1.00×10²-Ω, a 2.50-kΩ, and a 4.00-kΩ resistor connected in series? (b) In parallel?

a. (a) 6520 Ω (b) 570.4 Ω
b. (a) 6520 Ω (b) 1773.9 Ω
c. (a) 6500 Ω (b) 1775.0 Ω
d. (a) 570.4 Ω (b) 6520 Ω

User BigName
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1 Answer

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Final answer:

The total resistance for the resistors connected in series is 6600 ohms. For the parallel connection, the equivalent resistance is approximately 88.1 ohms.

Step-by-step explanation:

Calculating Series and Parallel Resistances

To find the total resistance of resistors connected in series, you add their resistances together. For the given resistors of 1.00 × 10² ohms (100 ohms), 2.50 kΩ (2500 ohms), and 4.00 kΩ (4000 ohms), the total series resistance is the sum of all these values, which is 100 ohms + 2500 ohms + 4000 ohms = 6600 ohms or 6.6 kΩ.

For parallel resistors, the equivalent resistance is found by using the formula 1/R_eq = 1/R_1 + 1/R_2 + 1/R_3. So for the resistors mentioned, the equivalent resistance in parallel would be calculated as follows: 1/R_eq = 1/100 + 1/2500 + 1/4000, which results in an equivalent resistance R_eq of approximately 88.1 ohms.

Therefore, the correct answer is: (a) 6600 ohms in series and (b) 88.1 ohms in parallel.

User Sino Raj
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