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Calculate the effective resistance across AB :

Calculate the effective resistance across AB. ICSE 2019 Physics Solved Question Paper.

Current Electricity

ICSE 2019

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Answer

The given circuit consists of three parts.

In the first part 5 Ω and 4 Ω are in series and if the equivalent resistance is given by Rs then,

Rs = 5 + 4 = 9 Ω

Then, in second part there is parallel combination of 3 Ω and Rs (9 Ω ), if equivalent resistance is Rp

1Rp=13+191Rp=3+191Rp=49Rp=94Rp=2.25\dfrac{1}{Rp} = \dfrac{1}{3} + \dfrac{1}{9} \\[0.5em] \Rightarrow \dfrac{1}{Rp} = \dfrac{3 + 1}{9} \\[0.5em] \Rightarrow \dfrac{1}{Rp} = \dfrac{4}{9} \\[0.5em] \Rightarrow Rp = \dfrac{9}{4} \\[0.5em] \Rightarrow R_p = 2.25 Ω

In the third part, 8 Ω and Rp ( 2.25 Ω ) are in series. Therefore, the equivalent resistance between AB i.e., RAB is

8 + 2.25 = 10.25 Ω .

Hence, Total resistance between A and B is 10.25 Ω

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