Solveeit Logo
Login

Question

Question: The equivalent resistance across AB is n $\Omega$. Find the value of n....

The equivalent resistance across AB is n Ω\Omega. Find the value of n.

Answer

1

Explanation

Solution

The circuit diagram shows three parallel paths between points A and B. Let's analyze each path:

  1. Direct Path: There is a 2Ω resistor connected directly between points A and B. Let's call this R1 = 2Ω.

  2. First Series Branch: There is a path from A, through an intermediate junction, to B. This path consists of two 2Ω resistors connected in series. The equivalent resistance of this series combination is R2 = 2Ω + 2Ω = 4Ω.

  3. Second Series Branch: Similarly, there is another path from A, through a different intermediate junction, to B. This path also consists of two 2Ω resistors connected in series. The equivalent resistance of this series combination is R3 = 2Ω + 2Ω = 4Ω.

Since these three paths are connected in parallel across points A and B, the equivalent resistance (R_eq) can be calculated using the formula for parallel resistors:

1Req=1R1+1R2+1R3\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

Substitute the values:

1Req=12 Ω+14 Ω+14 Ω\frac{1}{R_{eq}} = \frac{1}{2 \, \Omega} + \frac{1}{4 \, \Omega} + \frac{1}{4 \, \Omega}

To add these fractions, find a common denominator, which is 4:

1Req=24 Ω+14 Ω+14 Ω\frac{1}{R_{eq}} = \frac{2}{4 \, \Omega} + \frac{1}{4 \, \Omega} + \frac{1}{4 \, \Omega}

Add the numerators:

1Req=2+1+14 Ω\frac{1}{R_{eq}} = \frac{2 + 1 + 1}{4 \, \Omega}

1Req=44 Ω\frac{1}{R_{eq}} = \frac{4}{4 \, \Omega}

1Req=11 Ω\frac{1}{R_{eq}} = \frac{1}{1 \, \Omega}

Therefore, the equivalent resistance is:

Req=1 ΩR_{eq} = 1 \, \Omega

The problem states that the equivalent resistance across AB is n Ω. So, n = 1.