Question

Two resistors when connected in parallel have an equivalent of \(2 \Omega\) and when in series of \(9 \Omega\) The values of the two resistors are. (A) \(2 \Omega\) and \(9 \Omega\) (B) \(3 \Omega\) and \(6 \Omega\) (C) \(4 \Omega\) and \(5 \Omega\) (D) \(2 \Omega\) and \(7 \Omega\)

Short answer

There is no valid solution for the values of the two resistors based on the given information, as we obtain a negative resistance value which is not physically possible.

Step-by-step solution

Write the formula for equivalent resistance in parallel

The formula for the equivalent resistance (Rp) when two resistors are connected in parallel is: \( \frac{1}{Rp} = \frac{1}{R1} + \frac{1}{R2} \) Since we are given that the equivalent resistance when connected in parallel is 2Ω, we can substitute the value of Rp in the equation: \( \frac{1}{2} = \frac{1}{R1} + \frac{1}{R2} \)

Write the formula for equivalent resistance in series

The formula for the equivalent resistance (Rs) when two resistors are connected in series is: \( Rs = R1 + R2 \) Since we are given that the equivalent resistance when connected in series is 9Ω, we can substitute the value of Rs in the equation: \( 9 = R1 + R2 \)

Solve the equations simultaneously

Now we have two equations with two unknowns, R1 and R2. Let's solve these equations simultaneously to find the values of R1 and R2. From the series equation, we can write R2 in terms of R1: \( R2 = 9 - R1 \) Substitute this expression for R2 in the parallel equation: \( \frac{1}{2} = \frac{1}{R1} + \frac{1}{9 - R1} \) Now, we will solve for R1, then substitute the obtained value to get R2.

Solve for R1

To solve for R1, first, we'll make a common denominator: \( \frac{1}{2} = \frac{R1 (9 - R1)}{R1(9 - R1)} + \frac{R1}{R1(9 - R1)} \) Multiplying both sides by the common denominator, we can eliminate the fractions: \( R1(9 - R1) = \frac{1}{2}(R1^2 - 9R1 + R1^2) \) Now, solve the quadratic equation: \( 2R1^2 - 18R1 + 2R1^2 - R1^2 = 0 \) \( R1^2 - 18R1 = 0 \) \( R1(R1 - 18) = 0 \) So, R1 can be 0 Ω or 18 Ω, but since resistors cannot have 0 Ω, we go with R1 = 18 Ω.

Solve for R2

Now, we can substitute the value of R1 in the expression for R2 we wrote earlier: \( R2 = 9 - 18 \) \( R2 = -9 \) Since resistance cannot be negative, there appears to be an issue in the given information. Thus, we claim that there is no valid solution for the values of the two resistors based on the given information.

Key concepts

Understanding Parallel Circuits

Parallel circuits are a fundamental concept in electrical engineering. In a parallel circuit, components are connected across common points, creating multiple paths for the electrical current to flow. This means each component in the circuit operates independently. If one part fails, the rest can still function.

For resistors connected in parallel, the equivalent resistance is given by the formula:

• \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \dots \)

where \(R_p\) is the total equivalent resistance for resistors \(R_1, R_2, \ldots\).
This method is often used in practical applications to ensure devices remain operational even if one pathway is interrupted.

Exploring Series Circuits

Series circuits are another simple type of circuit configuration. In a series circuit, components are arranged in a single path. Thus, the current flows through each resistor sequentially. If one component fails, the entire circuit is broken.

The total resistance is simply the sum of individual resistances:

• \( R_s = R_1 + R_2 + \dots \)

where \(R_s\) is the equivalent resistance. This setup is straightforward but can limit total functionality if one component malfunctions.

Calculating Resistor Equations

Resistor equations are key in determining the total resistance in both parallel and series circuits. They help in analyzing circuits to ensure correct functionality and design.

For parallel circuits, the reciprocal rule enables calculating the equivalent resistance, while in series circuits, resistors directly add up. These calculations are vital for determining the correct specifications needed for circuit designs.

Resistor equations also assist in troubleshooting and optimizing electrical systems, providing clear insights into resistance management.

Approaching JEE Physics Problems

JEE Physics problems often include scenarios involving resistors in series and parallel configurations. These exercises assess comprehension of basic circuit laws and the ability to apply resistor equations.

To solve such problems, follow these strategies:

• Understand the circuit configuration, whether series or parallel.
• Identify given data, such as total resistance, individual resistances, or provided currents.
• Apply the appropriate formula: use the sum for series or reciprocal calculation for parallel.
• Ensure consistency in units to avoid calculation errors.

These steps encourage precision in solving circuit-related queries fundamental to achieving success in exams like the JEE.