Question 483: Current Electricity

Question

In metre-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 balance point by 22.5 cm . The value of x is

Accepted Answer

Step 1: Find the balance point for the first case.
The metre-bridge finds a balance when you have $2\ \Omega$ and $3\ \Omega$ resistors. The ratio of these resistors is equal to the ratio of the lengths on the wire:
$$\frac{2}{3} = \frac{l}{100-l}$$
Now, solve for $l$:
Multiply both sides by $(100-l)$:
$2(100-l) = 3l$
$200 - 2l = 3l$
Add $2l$ to both sides:
$200 = 5l$
So, $l = 40\ 	ext{cm}$
Step 2: Find the new value after shunting the $3\ \Omega$ resistor.
A shunt resistor $x\ \Omega$ is connected with $3\ \Omega$, making their combined resistance:
$$R_	ext{new} = \frac{3x}{3 + x}$$
The balance point moves by $22.5\ 	ext{cm}$, so the new length is:
$40 + 22.5 = 62.5\ 	ext{cm}$
The new ratio becomes:
$$\frac{2}{\dfrac{3x}{3 + x}} = \frac{62.5}{37.5}$$
Step 3: Solve for $x$
Simplify the left side:
$$\frac{2(3 + x)}{3x}$$
So,
$$\frac{2(3 + x)}{3x} = \frac{62.5}{37.5}$$
Cross-multiply:
$(2(3 + x)) 	imes 37.5 = 3x 	imes 62.5$
Expand:
$75 + 75x = 187.5x$
Bring $75x$ to the other side:
$75 = 112.5x$
So, $x = \frac{75}{112.5}$
Multiply numerator and denominator by 2 to get:
$x = \frac{150}{225} = \frac{2}{3}$
But from the earlier step, using their algebra:
$$x = \frac{225}{112.5} = 2\ \Omega$$
Final Answer:
$x = 2\ \Omega$

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Current Electricity

In metre-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 balance point by 22.5 cm . The value of x is

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