How can three resistors of resistances 2, 3, and 6 be connected to
give a total resistance of (a) 4 (b) 1
What is ſal the highest ſh the lowest total resistance that can be
Answers
Answer:
Please refer to the attachment also
Explanation:
(a) In order to obtain a total resistance of 4 Ω from three resistors of 2 Ω, 3 Ω and 6 Ω Firstly, connect the two resistors of 3Ω and 6Ω in parallel to get a total resistance of 2Ω which is less than the lowest individual resistance.
Then the net R from parallel connection is connected in series with the remaining 2 Q resistor to get a total resistance of 4 Q. This is because in series combination:
R=R1+ R2
R=2 + 2
R=4Ω
The arrangement of three resistors of 2,3Ω and 6Ω which gives a total resistance of 4Ω can now be represented as follows:
Thus, we can obtain a total resistance of 4Ω by connecting parallel combination of 3Ω and 6Ω resistors in series with 2Ω resistor.
(b) In order to obtain a total resistance of 1Ω from three resistors of 2Ω, 3Ω and 6Q, all the three resistors should be connected in parallel. This is because in parallel combination
Hope it helps ☺☺
:
Answer:
(a) In order to obtain a total resistance of 4 Ω from three resistors of 2 Ω, 3 Ω and 6 Ω Firstly, connect the two resistors of 3Ω and 6Ω in parallel to get a total resistance of 2Ω which is less than the lowest individual resistance.

Then the net R from parallel connection is connected in series with the remaining 2 Q resistor to get a total resistance of 4 Q. This is because in series combination:
R=R1+ R2
R=2 + 2
R=4Ω
The arrangement of three resistors of 2,3Ω and 6Ω which gives a total resistance of 4Ω can now be represented as follows:

Thus, we can obtain a total resistance of 4Ω by connecting parallel combination of 3Ω and 6Ω resistors in series with 2Ω resistor.
(b) In order to obtain a total resistance of 1Ω from three resistors of 2Ω, 3Ω and 6Q, all the three resistors should be connected in parallel. This is because in parallel combination:

Explanation:
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