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ICSE Class 10 Physics Question 7 of 31

Current Electricity — Question 8

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Question 8

You have three resistors of values 2 Ω, 3 Ω and 5 Ω . How will you join them so that the total resistance is less than 1 Ω ? Draw diagram and find the total resistance.

Answer

In order to get a total resistance less than 1 Ω, the three resistors of values 2 Ω, 3 Ω and 5 Ω, should be connected in parallel as shown in the diagram below:

You have three resistors of values 2 Ω, 3 Ω and 5 Ω . How will you join them so that the total resistance is less than 1 Ω ? Draw diagram and find the total resistance. Current Electricity, Concise Physics Solutions ICSE Class 10.

In parallel, if the equivalent resistance is RP, then

1RP=1R1+1R2+1R31RP=12+13+151RP=15+10+6301RP=3130⇒RP=0.97Ω\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \\[0.5em] \dfrac{1}{R_P} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{5} \\[0.5em] \dfrac{1}{R_P} = \dfrac{15 + 10+ 6}{30} \\[0.5em] \dfrac{1}{R_P} = \dfrac{31}{30} \\[0.5em] \Rightarrow R_P = 0.97 Ω

Hence, the three resistors should be connected in parallel and total resistance = 0.97 Ω