রোধ

Two unknown resistrance X and Y are connected to left and right gaps of a meter bridge and the balancing point is obtained at 80 cm from left. When a 10Ω\Omega resistance is connected in paralle to x, the balance point is 50 cm from left. The values of X and Y respectively are

হানি নাটস

Case 1

XY=8020=4 \Rightarrow \dfrac{X}{Y} = \dfrac{80}{20} = 4

X=4Y X = 4Y

Case 2

10X10+XY=5050=1 \dfrac{\dfrac{10X}{10+X}}{Y} = \dfrac{50}{50} = 1

1010+X×4=1 \Rightarrow \dfrac{10}{10+X} \times 4 = 1

10+X=40 10 + X = 40

X=30Ω X = 30 \Omega

Y=304=7.5Ω Y = \dfrac{30}{4} = 7.5 \Omega

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