The question describes a circuit with two resistors, $R_1$ and $R_2$, connected in parallel. We are given specific information about their physical properties: they are made of the same material and have the same thickness. This means they have the same resistivity ($\rho$) and the same cross-sectional area ($A$). We are also told that the length of $R_2$ is twice the length of $R_1$, which can be written as $L_2 = 2L_1$. We need to find the relationship between the total resistance ($R$) of the parallel combination and the individual resistances.
The resistance of a conductor is determined by its material, length, and cross-sectional area. The formula for resistance is:
$$R = \rho \frac{L}{A}$$
where:
Given that $R_1$ and $R_2$ are made of the same material and have the same thickness (same cross-sectional area $A$), their resistances can be written as:
We know that $L_2 = 2L_1$. Substituting this into the expression for $R_2$:
$$R_2 = \rho \frac{2L_1}{A}$$
We can rewrite this as:
$$R_2 = 2 \times \left( \rho \frac{L_1}{A} \right)$$
Since $R_1 = \rho \frac{L_1}{A}$, we can substitute $R_1$ into the equation for $R_2$:
$$R_2 = 2R_1$$
So, the resistance of $R_2$ is twice the resistance of $R_1$.
For two resistors connected in parallel, the total resistance $R$ is given by the formula:
$$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$$
Alternatively, the total resistance can be calculated using the product-sum formula:
$$R = \frac{R_1 R_2}{R_1 + R_2}$$
We found that $R_2 = 2R_1$. Now, substitute this relationship into the parallel resistance formula $R = \frac{R_1 R_2}{R_1 + R_2}$:
$$R = \frac{R_1 (2R_1)}{R_1 + 2R_1}$$
Simplify the expression:
$$R = \frac{2R_1^2}{3R_1}$$
Cancel out one $R_1$ term from the numerator and the denominator:
$$R = \frac{2R_1}{3}$$
To match the format of the options, we can rearrange this equation by multiplying both sides by 3:
$$3R = 2R_1$$
Let's compare our derived relationship $3R = 2R_1$ with the given options:
Our result $3R = 2R_1$ matches Option 1.
| Concept | Description | Formula |
|---|---|---|
| Resistance (R) | Opposition to electric current flow. | $R = \rho \frac{L}{A}$ |
| Resistivity ($\rho$) | Intrinsic property of a material indicating its resistance. | Unit: Ohm-meter ($\Omega \cdot m$) |
| Resistance in Series | Total resistance is the sum of individual resistances. | $R_{total} = R_1 + R_2 + \dots + R_n$ |
| Resistance in Parallel | Reciprocal of total resistance is the sum of reciprocals of individual resistances. | $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}$ |
Understanding the factors affecting resistance and how components behave in series versus parallel is crucial for circuit analysis.
Three equal resistors are connected in parallel configuration in a closed electrical circuit. Then the total resistance in the circuit becomes
A metallic wire having a resistance of 20Ω is cut into two equal parts in length. These parts are then connected in parallel. The resistance of this parallel combination is equal to
Consider two resistors, $R_1$ and $R_2$, connected in series to a DC voltage source. Which of the following statements accurately describes the distribution of current and voltage across these resistors?
A cell of negligible resistance and e.m.f 2 volt is connected to series combination of 2 ohm, 3 ohm and 5 ohm. The potential difference across the 3 ohm resistance is:
Two bulbs A, of (100w, 100v), and B of (60 w, 100v) are connected in series and across the series combination 200 v is applied. Which bulb will be fused?
3 resistors of 3 ohm each connected in series. What is the mean values of resistors?
The equivalent resistance of the resistances (two) joined in parallel is 6/5 Ω. When one of the resistance wire is broken, the effective resistance becomes 2Ω. The resistance of the wire that got broken was :