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a 51-kω resistor and a 39-kω resistor are in parallel, and the pair is in series with a 22-kω resistor. What is the resistance of the combination?

User Kalid
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Final answer:

To find the total resistance of the combination, first calculate the parallel equivalent resistance, then add the series resistance. The combined resistance is approximately 44.3 kΩ.

Step-by-step explanation:

The question asks for the total resistance of a circuit containing a 51-kΩ resistor and a 39-kΩ resistor in parallel, with this pair in series with a 22-kΩ resistor. To find the total resistance, we calculate the equivalent resistance of the parallel combination before adding it to the resistance of the series resistor. The equivalent resistance for resistors in parallel is given by 1/Rp = 1/R1 + 1/R2. For the two resistors in parallel:

1/Rp = 1/51 + 1/39

1/Rp = 39/2007 + 51/2007

1/Rp = 90/2007

Rp = 2007/90 kΩ = 22.3 kΩ (approx)

Now, we add this equivalent parallel resistance to the series resistance:

Total Resistance = Rp + 22 kΩ

Total Resistance = 22.3 kΩ + 22 kΩ = 44.3 kΩ (approx)

Therefore, the total resistance of the combination is approximately 44.3 kΩ.

User Ragebiswas
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