Calculate the effective resistance across AB :

ICSE Class 10 Physics PYQ · 2019 Paper · Section I

4(a)2 Marks · 2019 · Section I · 2019 Paper

Calculate the effective resistance across AB :

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

$$\begin{aligned}\frac{1}{R_p} = \frac{1}{3} + \frac{1}{9} \\[0.5em] \Rightarrow \frac{1}{R_p} = \frac{3 + 1}{9} \\[0.5em] \Rightarrow \frac{1}{R_p} = \frac{4}{9} \\[0.5em] \Rightarrow R_p = \frac{9}{4} \\[0.5em] \Rightarrow R_p = 2.25 Ω\end{aligned}$$

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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