Three resistors of 6 Ω, 3 Ω and 2 Ω are connected together so that the total resistance is greater than 6 Ω but less than 8 Ω. Draw a diagram to show this arrangement and calculate its total resistance.
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If you connect the 3Ω and 2Ω resistors in parallel, their equivalent resistance would be: 3*2/5 = 6/5 = 1.2Ω
Then connect the 6Ω resistor to it in series. Then the final total resistance would be:
6Ω + 1.2Ω = 7.2Ω which is greater than 6Ω and less than 8Ω
Here's the diagram:
Attachments:
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