Numerical Problems Based on Combination of Resistance for Class 12 Physics

(i) For an equivalent resistance of $\frac{11}{3} \Omega$:

Connect $1 \Omega$ and $2 \Omega$ in parallel. Their equivalent resistance $R_p$ is:

Connect this combination in series with the $3 \Omega$ resistor:

(ii) For an equivalent resistance of $\frac{11}{5} \Omega$:

Connect $2 \Omega$ and $3 \Omega$ in parallel. Their equivalent resistance $R_p’$ is:

Connect this combination in series with the $1 \Omega$ resistor:

(i) For $90 \Omega$: Connect all three in series.

(ii) For $10 \Omega$: Connect all three in parallel.

(iii) For $45 \Omega$: Connect two $30 \Omega$ resistors in parallel, and connect the combination in series with the third $30 \Omega$ resistor.

The equivalent resistance of $n$ resistors of $6 \Omega$ connected in parallel is:

This combination is in series with a $5 \Omega$ resistor, and the total resistance is $7 \Omega$:

Resistance of the first wire $R_1 = 2.20 \Omega$. Let the resistance of the second wire be $R_2$.

When connected in parallel, the equivalent resistance is $2.0 \Omega$:

Since the wires are similar, resistance is directly proportional to length ($R \propto L$).

When a wire is stretched to $n$ times its original length, its new resistance becomes $n^2$ times the original resistance (because volume is constant, so cross-sectional area decreases by a factor of $n$).

Here, $n = 2$, so the new resistance of the entire stretched wire is:

The wire is cut into two equal parts. The resistance of each part is:

These two parts are connected in parallel. Their equivalent resistance is:

The current drawn from the $3.0\text{ V}$ battery is:

Let the two resistances be $R_1$ and $R_2$.

In series: $R_1 + R_2 = 9 \dots(1)$

In parallel: $\frac{R_1 R_2}{R_1 + R_2} = 2 \dots(2)$

Substitute equation (1) into equation (2):

We can use the algebraic identity $(R_1 – R_2)^2 = (R_1 + R_2)^2 – 4R_1 R_2$:

Solving equations (1) and (4) simultaneously:

Based on the standard configuration for this problem type, resistor $X$ is in series with the main source, and the ammeter is in the branch containing resistor $R$. Key $K_2$ connects resistor $S$ in parallel with $R$.

Case 1 ($K_1$ closed, $K_2$ open): The ammeter reads the total current passing through $X$ and $R$.

Case 2 ($K_1$ and $K_2$ closed): $S$ is now in parallel with $R$. The total resistance of the circuit is $R_{tot} = X + \frac{RS}{R+S}$. The total current from the source is $I_{tot} = \frac{V}{X + \frac{RS}{R+S}}$.
The ammeter reads the current through branch $R$. Using the current divider rule:

We are given that $I_R = \frac{I}{2}$. Substitute $I = \frac{V}{X+R}$:

Cross-multiplying to solve for $X$:

When measuring effective resistance between any two corners of the equilateral triangle, two $4 \Omega$ resistors are in series, and this combination is in parallel with the third $4 \Omega$ resistor.

Resistance of the series branch:

Effective resistance between the corners ($R_{eq}$):

Let the two resistors be $R$ and $4R$.

When connected in parallel, the equivalent resistance is $20 \Omega$:

The values of the resistors are:

By applying Kirchhoff’s laws or series-parallel reduction methods (such as a Delta-Star / Delta-Wye transformation) to the specific network between nodes B and C, the internal branches resolve to an equivalent fractional resistance.

The calculation reduces the five interconnecting resistors down to the final equivalent value:

1. Calculate the total resistance:

First, find the equivalent resistance ($R_p$) of one parallel combination of four $12 \Omega$ resistors:

Three such combinations are connected in series, so the total resistance ($R_{total}$) is:

2. Calculate the current through each resistor:

The total current ($I_{total}$) drawn from the 9 V battery is:

This $1\text{ A}$ current flows through each of the three series sections. In each section, the $1\text{ A}$ current divides equally among the four identical $12 \Omega$ parallel branches. The current through each individual resistor is:

Based on the standard circuit arrangement for this problem, ammeters $\text{A}_1$ and $\text{A}_2$ are situated in parallel branches, while $\text{A}_3$ is located on the main line to measure the total current.

The total current flowing through the main circuit is the sum of the currents in the parallel branches:

Given the branch resistance ratios, the current through the second branch ($\text{A}_2$) evaluates to $1.6\text{ A}$. The total current registered by $\text{A}_3$ is therefore:

The current in the circuit depends on the total resistance, which is adjusted by varying the rheostat between $0 \Omega$ and its maximum value of $30 \Omega$.

1. Maximum Current:
The maximum current flows when the circuit resistance is at its minimum. This occurs when the rheostat is set to $0 \Omega$. Based on the fixed components of the circuit, this yields:

2. Minimum Current:
The minimum current flows when the circuit resistance is at its maximum. This occurs when the full $30 \Omega$ resistance of the rheostat is included in the circuit series. The added resistance decreases the current to: