Solved 2019 Question Paper ICSE Class 10 Physics — Question 16

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 R_s then,

R_s = 5 + 4 = 9 Ω

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

$\dfrac{1}{R_p} = \dfrac{1}{3} + \dfrac{1}{9} \\[0.5em] \Rightarrow \dfrac{1}{R_p} = \dfrac{3 + 1}{9} \\[0.5em] \Rightarrow \dfrac{1}{R_p} = \dfrac{4}{9} \\[0.5em] \Rightarrow R_p = \dfrac{9}{4} \\[0.5em] \Rightarrow R_p = 2.25 Ω$

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

8 + 2.25 = 10.25 Ω .

Hence, Total resistance between A and B is 10.25 Ω