Question

There are three resistors having resistances R1, R2 and R3. R1 < R2 < R3. Which of the following mathematical formula will give the value of equivalent resistance which is more than the lowest individual resistance in the circuit but lesser than the highest individual resistance in the circuit?
a. Connect parallel combination of R1 and R2 in series with R3.

b. Connect parallel combination of R2 and R3 in series with R1.

c. Connect parallel combination of R1 and R3 in series with R2.

d. Connect all the three resistors in parallel.

Answer 1

Answer:

b. connect parallel combination of R2 and r3 in series with r1

Explanation:

connecting R2 and R3 in parallel will give resistance(R23) < R2 which will be <R3.

connecting that in series with R1 will give resistance more than R1 but as R1 and R23 is less than R3 so equivalent resistance will be more than R1 but less than R3.

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thankx

Answer 2

Given : Three resistors having resistances R1, R2 and R3

             R1 < R2 < R3

To Find : Equivalent resistance which is more than the lowest individual         resistance in the circuit but lesser than the highest individual resistance in the circuit

Solution :

Resistance in series

$\[{R_{TOTAL}} = R1 + R2 + R3.\]$

Parallel circuit,

Connecting R2 and R3 in parallel

• It will give resistance($\[{R_{23}}\]$) < R2 which will be <R3.

• Connecting that in series with R1 will give resistance more than R1 but as R1 and $\[{R_{23}}\]$ is less than R3 so equivalent resistance will be more than R1 but less than R3.

Hence, equivalent resistance which is more than the lowest individual         resistance in the circuit but lesser than the highest individual resistance in the circuit is (b. Connect parallel combination of R2 and R3 in series with R1.)