Two resistors R1 and R2 are connected in parallel, R2 being greater than R1. The combined resistance is
less than R1
greater than R2
the sum of R1 and R2
the difference of R2 and R1
greater thanR1 but less than R2
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What is this now??? Resistors in parallel.. Effective resistance is 1/R=1/r+1/r hence A is the answer pleaseeee

e connected in parallel, the combined resistance
is given by the formula:
1/Rโ=1/R1โ+1/R2โ
This results in a resistance that is less than the smallest individual resistor. So, if R2 is greater than R1, the combined resistance will be less than R1. This is because the parallel connection provides multiple paths for the current to flow, effectively reducing the total resistance.

The correct answer is E) greater than R1 but less than R2.
When two resistors are connected in parallel, the total resistance decreases. The combined resistance can be calculated using the following formula:
1/total resistance = 1/R1 + 1/R2
Rearranging this formula gives:
total resistance = R1R2 / (R1 + R2)
Since R2 is greater than R1, the denominator R1 + R2 is greater than R1 alone but less than R2 alone. Therefore, the total resistance is greater than R1 but less than R2.
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