Current Electricity - Wheatstone Bridge

In a Wheatstone bridge, the four arms have resistances $P = 100\, \Omega$, $Q = 10\, \Omega$, $R = 50\, \Omega$, and $S$ is unknown. If the bridge is balanced, calculate the value of $S$.

Given: $P = 100\, \Omega$, $Q = 10\, \Omega$, $R = 50\, \Omega$. Using the balance condition $\frac{P}{Q} = \frac{R}{S}$, we have: $\frac{100}{10} = \frac{50}{S} \implies 10 = \frac{50}{S} \implies S = \frac{50}{10} = 5\, \Omega$.

The balanced Wheatstone bridge condition states that the product of opposite arm resistances is equal, or the ratios of adjacent arms are equal. Here, we solve for the unknown arm $S$ by substituting the known values into the ratio formula.

In a Meter Bridge experiment, the null point is found at a distance of $40\text{ cm}$ from end $A$ when a resistance $R = 12\, \Omega$ is connected in the left gap. Find the value of the unknown resistance $S$ in the right gap.

Given: $l = 40\text{ cm}$, $R = 12\, \Omega$. The length of the remaining wire is $(100 - l) = 100 - 40 = 60\text{ cm}$. Using the formula $S = \frac{100 - l}{l} \times R$: $S = \frac{60}{40} \times 12 = \frac{3}{2} \times 12 = 18\, \Omega$.

The Meter Bridge works on the principle of the Wheatstone bridge. The ratio of the resistances in the two gaps equals the ratio of the lengths of the two segments of the wire at the balance point.

In the given Wheatstone bridge circuit, the resistances in the arms are $P = 15\, \Omega$, $Q = 30\, \Omega$, and $R = 10\, \Omega$. A galvanometer is connected between points $B$ and $D$. Determine the value of the unknown resistance $S$ required to balance the bridge and find the current through the galvanometer when the bridge is balanced.

For a Wheatstone bridge to be balanced, the condition is given by: $\frac{P}{Q} = \frac{R}{S}$

Substituting the given values: $\frac{15}{30} = \frac{10}{S}$

$\frac{1}{2} = \frac{10}{S}$

$S = 10 \times 2 = 20\, \Omega$

When the bridge is balanced, the potential at point $B$ is equal to the potential at point $D$ ($V_B = V_D$). Therefore, the potential difference across the galvanometer is zero, and the current through it is: $I_g = 0\, \text{A}$

The balancing condition ensures that no current flows through the central arm (galvanometer). This happens when the ratio of resistances in the upper arms equals the ratio of resistances in the lower arms.

A Meter Bridge is used to find an unknown resistance $S$. When a standard resistor $R = 5\, \Omega$ is placed in the left gap, the null point is obtained at $l = 40\text{ cm}$ from the left end $A$. Calculate the value of the unknown resistance $S$ in the right gap. If the positions of $R$ and $S$ are interchanged, what will be the new balancing length from end $A$?

For a Meter Bridge, the balancing condition is: $\frac{R}{S} = \frac{l}{100 - l}$

Given $R = 5\, \Omega$ and $l = 40\text{ cm}$: $\frac{5}{S} = \frac{40}{100 - 40}$

$\frac{5}{S} = \frac{40}{60} = \frac{2}{3}$

$S = \frac{5 \times 3}{2} = 7.5\, \Omega$

When $R$ and $S$ are interchanged, let the new balance length be $l'$. The condition becomes: $\frac{S}{R} = \frac{l'}{100 - l'}$

$\frac{7.5}{5} = \frac{l'}{100 - l'}$

$1.5 = \frac{l'}{100 - l'}$

$1.5(100 - l') = l'$

$150 - 1.5l' = l'$

$150 = 2.5l'$

$l' = \frac{150}{2.5} = 60\text{ cm}$

The Meter Bridge works on the principle of the Wheatstone bridge. The resistance of the wire is proportional to its length, so the ratio of resistances equals the ratio of the lengths of the two segments of the wire.