A resistor of 4kΩ with tolerance 10% is connected in parallel with a resistor of 6 kΩ with tolerance 10%. The tolerance of the parallel combination is nearly ______?
Answer: 10%
Explanation
The equivalent resistance (R_eq) of the parallel combination:
R_eq = (4 kΩ × 6 kΩ) / (4 kΩ + 6 kΩ) = 2.4 kΩ
The tolerance calculation:
Since both resistors have the same relative tolerance (10%), the resultant parallel combination will also have a tolerance close to 10%.
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