Question 536: Current Electricity

Question

In a meter bridge experiment, the balance point is obtained if the gaps are closed by $2 \Omega$ and $3 \Omega$. A shunt of $\mathrm{X} \Omega$ is added to $3 \Omega$ resistor to shift the null point by 22.5 cm. The value of ' $x$ ' is

Accepted Answer

Case 1: Initial Condition
The balance equation is given by:
$ \frac{2}{3} = \frac{l}{100 - l} $
Solving for $l$:
$ 200 - 2l = 3l \ 5l = 200 \ l = 40 \, 	ext{cm} $
Case 2: With Shunt
Now, with the shunt, the resistance changes and the length becomes $40 + 22.5$ cm, leading to the equation:
$ \frac{2}{\frac{3X}{3+X}} = \frac{40 + 22.5}{100 - (40 + 22.5)} $
Substitute and simplify the equation:
$ \frac{2(3+X)}{3X} = \frac{62.5}{37.5} $
Solving for $X$:
$ (6 + 2X) 	imes 37.5 = 3X 	imes 62.5 $
Expanding and simplifying gives:
$ 225 + 75X = 187.5X \ 187.5X - 75X = 225 \ 112.5X = 225 \ X = \frac{225}{112.5} = 2 \, \Omega $
Thus, $X$ is $2 \, \Omega$.

Answer

$1 \Omega$

Answer

$2 \Omega$

Answer

$3 \Omega$

Answer

$4 \Omega$

Current Electricity

In a meter bridge experiment, the balance point is obtained if the gaps are closed by 2 \Omega and 3 \Omega. A shunt of \mathrm{X} \Omega is added to 3 \Omega resistor to shift the null point by 22.5 cm. The value of ' x ' is

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