Key Concept: Combination of resistors in series and parallel
a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let's analyze the assertion and reason step by step.
1. First, consider the parallel combination of two resistors $R_1$ and $R_2$. The equivalent resistance $R_p$ of these two resistors in parallel is given by:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$
Simplifying this, we get:
$R_p = \frac{R_1 R_2}{R_1 + R_2}$
Here, $R_p$ is always less than both $R_1$ and $R_2$ because the denominator $(R_1 + R_2)$ is greater than either $R_1$ or $R_2$. This supports the reason statement.
2. Now, this parallel combination is connected in series with the third resistor $R_3$. The total equivalent resistance $R_{eq}$ of the circuit is:
$R_{eq} = R_p + R_3 = \frac{R_1 R_2}{R_1 + R_2} + R_3$
Since $R_p < R_1$ and $R_p < R_2$, but we are adding $R_3$ to $R_p$, it is possible that $R_{eq}$ could be greater than at least one of the resistors in the circuit if $R_3$ is large enough. However, the assertion claims that $R_{eq}$ will always be greater than the smallest individual resistor, which may not necessarily be true if the smallest resistor is $R_3$.
For example, let $R_1 = 4 \Omega$, $R_2 = 4 \Omega$, and $R_3 = 1 \Omega$. Then,
$R_p = \frac{4 \times 4}{4 + 4} = 2 \Omega$
And $R_{eq} = 2 + 1 = 3 \Omega$. Here, $3 \Omega$ is greater than the smallest resistor $R_3 = 1 \Omega$, so the assertion holds.
But if $R_1 = 1 \Omega$, $R_2 = 1 \Omega$, and $R_3 = 4 \Omega$, then:
$R_p = \frac{1 \times 1}{1 + 1} = 0.5 \Omega$
And $R_{eq} = 0.5 + 4 = 4.5 \Omega$. Here, $4.5 \Omega$ is greater than the smallest resistor $R_1 = 1 \Omega$, so the assertion still holds.
From these examples, it appears that the assertion is true under most conditions where the smallest resistor is part of the parallel combination. The reason correctly explains why the parallel combination results in a lower resistance, supporting the assertion when the smallest resistor is in the parallel branch. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.
Your Answer is correct.
a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Let's analyze the assertion and reason step by step.
1. First, consider the parallel combination of two resistors $R_1$ and $R_2$. The equivalent resistance $R_p$ of these two resistors in parallel is given by:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$
Simplifying this, we get:
$R_p = \frac{R_1 R_2}{R_1 + R_2}$
Here, $R_p$ is always less than both $R_1$ and $R_2$ because the denominator $(R_1 + R_2)$ is greater than either $R_1$ or $R_2$. This supports the reason statement.
2. Now, this parallel combination is connected in series with the third resistor $R_3$. The total equivalent resistance $R_{eq}$ of the circuit is:
$R_{eq} = R_p + R_3 = \frac{R_1 R_2}{R_1 + R_2} + R_3$
Since $R_p < R_1$ and $R_p < R_2$, but we are adding $R_3$ to $R_p$, it is possible that $R_{eq}$ could be greater than at least one of the resistors in the circuit if $R_3$ is large enough. However, the assertion claims that $R_{eq}$ will always be greater than the smallest individual resistor, which may not necessarily be true if the smallest resistor is $R_3$.
For example, let $R_1 = 4 \Omega$, $R_2 = 4 \Omega$, and $R_3 = 1 \Omega$. Then,
$R_p = \frac{4 \times 4}{4 + 4} = 2 \Omega$
And $R_{eq} = 2 + 1 = 3 \Omega$. Here, $3 \Omega$ is greater than the smallest resistor $R_3 = 1 \Omega$, so the assertion holds.
But if $R_1 = 1 \Omega$, $R_2 = 1 \Omega$, and $R_3 = 4 \Omega$, then:
$R_p = \frac{1 \times 1}{1 + 1} = 0.5 \Omega$
And $R_{eq} = 0.5 + 4 = 4.5 \Omega$. Here, $4.5 \Omega$ is greater than the smallest resistor $R_1 = 1 \Omega$, so the assertion still holds.
From these examples, it appears that the assertion is true under most conditions where the smallest resistor is part of the parallel combination. The reason correctly explains why the parallel combination results in a lower resistance, supporting the assertion when the smallest resistor is in the parallel branch. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.