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Question

Two resistors of R Ω and 15 Ω are connected in parallel to get an effective resistance of 12 Ω. Find R.

This question was previously asked in
RRB ALP 2018 CBT 2 Fitter Question Paper (21-Jan-2019) (Shift 3)
The correct answer is

60

Understanding Resistors in Parallel Circuits

In this problem, we are given two resistors connected in parallel. One resistor has a resistance of \(R \, \Omega\), and the other has a resistance of \(15 \, \Omega\). We are told that the combined or effective resistance of this parallel combination is \(12 \, \Omega\). Our goal is to find the value of the unknown resistance \(R\).

Parallel Resistance Calculation

When resistors are connected in parallel, the reciprocal of the effective resistance (\(R_{eq}\)) is equal to the sum of the reciprocals of the individual resistances. The formula for two resistors, \(R_1\) and \(R_2\), connected in parallel is:

\(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\)

An alternative formula, often useful for just two resistors, is:

\(R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}\)

Step-by-Step Solution to Find R

Let's use the second formula with the given values:

  • \(R_1 = R\) (the unknown resistance)
  • \(R_2 = 15 \, \Omega\) (the known resistance)
  • \(R_{eq} = 12 \, \Omega\) (the effective resistance)

Substitute these values into the formula:

\(12 \, \Omega = \frac{R \times 15 \, \Omega}{R + 15 \, \Omega}\)

Now, we need to solve this equation for \(R\):

  1. Multiply both sides by \((R + 15)\) to remove the denominator:

    \(12 \times (R + 15) = 15R\)

  2. Distribute the 12 on the left side:

    \(12R + (12 \times 15) = 15R\)

    \(12R + 180 = 15R\)

  3. Subtract \(12R\) from both sides of the equation to isolate the terms with \(R\) on one side:

    \(180 = 15R - 12R\)

    \(180 = 3R\)

  4. Divide both sides by 3 to find the value of \(R\):

    \(R = \frac{180}{3}\)

    \(R = 60 \, \Omega\)

So, the value of the unknown resistor \(R\) is \(60 \, \Omega\).

Verifying the Result

Let's quickly check if a \(60 \, \Omega\) resistor in parallel with a \(15 \, \Omega\) resistor gives an effective resistance of \(12 \, \Omega\):

\(\frac{1}{R_{eq}} = \frac{1}{60} + \frac{1}{15}\)

Find a common denominator, which is 60:

\(\frac{1}{R_{eq}} = \frac{1}{60} + \frac{4}{60}\)

\(\frac{1}{R_{eq}} = \frac{1 + 4}{60}\)

\(\frac{1}{R_{eq}} = \frac{5}{60}\)

Simplify the fraction:

\(\frac{1}{R_{eq}} = \frac{1}{12}\)

Taking the reciprocal of both sides:

\(R_{eq} = 12 \, \Omega\)

This matches the given effective resistance, confirming our calculation is correct.

Revision Table: Key Concepts

Concept Description Formula (for two resistors)
Resistors in Parallel Components are connected across the same two points, providing alternative paths for current. \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}\) or \(R_{eq} = \frac{R_1 R_2}{R_1 + R_2}\)
Effective Resistance (\(R_{eq}\)) The total resistance of a circuit or part of a circuit. In parallel, \(R_{eq}\) is always less than the smallest individual resistance. Calculated using the parallel resistance formula.

Additional Information: Parallel vs. Series Circuits

Understanding the difference between parallel and series connections is crucial in circuit analysis. Here’s a brief comparison:

  • Series Connection: Components are connected end-to-end, forming a single path for the current. The total resistance is the sum of individual resistances (\(R_{eq} = R_1 + R_2 + R_3 + ...\)). The current is the same through all components.
  • Parallel Connection: Components are connected across the same two points, providing multiple paths for the current. The reciprocal of the total resistance is the sum of the reciprocals of individual resistances. The voltage is the same across all components in parallel.

Parallel connections are commonly used in household wiring so that each appliance receives the full voltage and can be operated independently.

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