Two resistors 2Ω$2\Omega$ and 3Ω$3\Omega$ are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire XY. When an unknown resistor is connected in parallel with 3Ω$3\Omega$ resistor, the null point is shifted by 22.5 cm toward Y. The resistance of unknown resistor is ____ Ω$\Omega$.

Core Logic

In steady state, the capacitor acts as an open circuit. No current flows through the branch containing the 10$10\mu\text{F}$ capacitor.

After removing the capacitor branch, we simplify the network of resistors.

Step 1: Equivalent Resistance

By further simplifying the parallel and series combinations, we determine the main current and the current branching into the ammeter.

Step 2: Ammeter Reading

The current of 2A$2\mathrm{A}$ splits equally through the symmetrical 8Ω$8\Omega$ branches. The ammeter reading is 1A$1\mathrm{A}$.

Pattern Recognition

Steady state DC + Capacitor = Open circuit. Always delete the capacitor branch first before calculating equivalent resistance.

Chapter Mix

Class 12 Physics: Current Electricity Class 12 Physics: Electrostatics

Core Logic

By applying the principle of symmetry to the regular hexagonal resistor network, the circuit can be simplified across the axis of symmetry.

Step 1: Network Reduction

Using equipotential nodes along the perpendicular bisector, we can resolve the network into parallel and series components.

The equivalent resistance for each half is determined as follows:

Step 2: Final Equivalent Resistance

Combining the simplified branches:

Pattern Recognition

For regular polygonal networks with central connections, always search for a line of equipotential symmetry perpendicular to the line joining the input and output nodes. Nodes on this axis can be detached or merged.

Chapter Mix

Class 12 Physics: Current Electricity

Related Formula

Meter bridge balance condition:

Core Logic

In Case I:

Step 1: Calculate initial null point

Solving Equation (1):

Step 2: Apply Case II conditions

In Case II, the right gap resistance becomes a parallel combination: R' = (3x)/(3+x)$R' = \frac{3x}{3+x}$. The new null point is shifted 10 cm$10 \text{ cm}$ to the right, meaning the new balancing length is ' = 40 + 10 = 50 cm$\ell' = 40 + 10 = 50 \text{ cm}$.

Thus, R' = 2 Ω$R' = 2 \, \Omega$.

Step 3: Solve for x

3x = 6 + 2x$3x = 6 + 2x$ x = 6 Ω$x = 6 \, \Omega$

Pattern Recognition

If the null point falls exactly at 50 cm$50 \text{ cm}$, the two gap resistances must be identical. Recognizing this immediately yields Rparallel = 2$R_{\text{parallel}} = 2$ without solving proportional fractions.

Chapter Mix

Class 12 Physics: Current Electricity

Related Formula

For a cell of emf E$E$ and internal resistance r$r$ shunted by a resistance R$R$, the terminal voltage is:

In a potentiometer, the terminal voltage is proportional to the balancing length l$l$: V = K · l$V = K \cdot l$

Core Logic

Equate the expressions for terminal voltage for both cases to their respective balancing lengths. Let E$E$ be the emf and r$r$ be the internal resistance of the cell.

Step 1: First Condition Setup

When shunted with 4 Ω$4 \,\Omega$:

Step 2: Second Condition Setup

When shunted with 12 Ω$12 \,\Omega$:

Step 3: Solving for Internal Resistance

Divide equation (i) by equation (ii):

r + 12 = 2r + 8$r + 12 = 2r + 8$ r = 4 Ω$r = 4 \,\Omega$

Pattern Recognition

Standard potentiometer problem. Ratio of balancing lengths maps directly to the ratio of terminal voltages. (l₁)/(l₂) = (R₁(r+R₂))/(R₂(r+R₁))$\frac{l_1}{l_2} = \frac{R_1(r+R_2)}{R_2(r+R_1)}$. Plugging the values simplifies it instantly.

Chapter Mix

Class 12 Physics: Current Electricity