Question

The effective resistance of a n number of resistors connected in parallel in \(\mathrm{x} \mathrm{ohm}\). When one of the resistors is removed, the effective resistance becomes y ohm. The resistance of the resistor that is removed is.... (A) \(\\{(\mathrm{xy}) /(\mathrm{x}+\mathrm{y})\\}\) (B) \(\\{(\mathrm{xy}) /(\mathrm{y}-\mathrm{x})\\}\) (C) \((\mathrm{y}-\mathrm{x})\) (D) \(\sqrt{x y}\)

Short answer

The resistance of the resistor that was removed is (B) \(\frac{xy}{y-x}\) ohm.

Step-by-step solution

Understand the Problem

There are n resistors connected in parallel with a total resistance of x ohms. If one resistor is removed, the total resistance becomes y ohms. The aim is to find the resistance of that removed resistor. This requires understanding the formula for total resistance in a parallel circuit.

Recall the Formula for Resistors in Parallel

The formula for n number of resistors (R1, R2, R3,..., Rn) connected in parallel can be written as: \[ \frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} + \ldots + \frac{1}{Rn} \] Where R is the total resistance. Now, before removing the resistor, let's say the resistance is R1 and the resistance of all other resistors in parallel is R2. So, R (total resistance before removing the resistor) is x and is calculated as: \[ \frac{1}{x} = \frac{1}{R1} + \frac{1}{R2} \] After removing the resistor, the resistance is y and is calculated as: \[ \frac{1}{y} = \frac{1}{R2} \]

Solve for R1

Now, to find R1, firstly, let's express \(\frac{1}{R1}\) in terms of x and y using the formulas from second step: \[ \frac{1}{R1} = \frac{1}{x} - \frac{1}{y} \] The value of resistor R1 (the resistor that has been removed) is just reciprocal of the above expression: \[ R1= \frac{1}{\frac{1}{R1}} \] This simplifies to: \[ R1= \frac{xy}{y-x} \]

Check for the Correct Option

Looking at the available options, we can see that option (B) corresponds to our calculated Result R1. Therefore, the resistance of the resistor that was removed is (B) \(\frac{xy}{y-x}\) ohm.

Key concepts

Effective Resistance

Effective resistance is an important concept in understanding how resistors work in a circuit, especially when they are arranged in parallel. When multiple resistors are connected in parallel, the total or effective resistance is less than the resistance of the smallest individual resistor. This is because each resistor provides an additional path for current, reducing the overall resistance.

In parallel circuits, the effective resistance can be thought of as the combined resistance that would allow the same amount of current to flow as the individual resistors do collectively. The formula for finding effective resistance (R ) of resistors in parallel is:

• If two resistors, say R_1 and R_2 , are in parallel, the effective resistance R can be calculated using: \[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \]

With this formula, you can find the effective resistance of any number of resistors connected in parallel by continuing the pattern of summing the reciprocals of each resistor's resistance.

Circuit Analysis

Analyzing circuits with parallel resistors is a foundational skill in electronics. It involves understanding how currents and voltages behave in a parallel arrangement. In a parallel circuit, all components share the same voltage across them, but the current can differ through each branch.

When analyzing such a circuit, it's crucial to:

• Identify all resistors connected in parallel.
• Use the parallel resistance formula to find the total resistance.
• Evaluate how removing or adding a resistor affects the total resistance.

In the given problem, when one resistor is removed and the effective resistance changes from x to y ohms, circuit analysis helps identify the resistance of the removed resistor. This is achieved by recalculating the effective resistance without the removed resistor and comparing it to the original configuration.

Resistor Formulas

Resistor formulas are key to solving problems in electric circuits, particularly when they involve resistors connected in parallel. In addition to the parallel resistance formula, being comfortable with basic algebraic manipulation and reciprocals is essential.

The initial formula for R1 removed from the circuit is derived from:

• Before removal: \[ \frac{1}{x} = \frac{1}{R1} + \frac{1}{R2} \]
• After removal: \[ \frac{1}{y} = \frac{1}{R2} \]

Using these, the resistance of the removed resistor can be found as:

• Subtracting the two equations to isolate \( \frac{1}{R1} \): \[ \frac{1}{R1} = \frac{1}{x} - \frac{1}{y} \]
• The reciprocal of \( \frac{1}{R1} \) gives R1 's resistance: \[ R1 = \frac{xy}{y-x} \]

This manipulation demonstrates how effective resistance changes when resistors are added or removed, guiding decisions in circuit design and troubleshooting.