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Chapter 12: Problem 1720

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

Expert verified
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

01

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: \( \frac{1}{2} = \frac{1}{R1} + \frac{1}{R2} \)
02

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: \( 9 = R1 + R2 \)
03

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.
04

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.
05

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.

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Most popular questions from this chapter

The resistance of a copper coil is \(4.64 \Omega\) at \(40^{\circ} \mathrm{C}\) and \(5.6 \Omega\) at \(100^{\circ} \mathrm{C}\) Its resistance at $0^{\circ} \mathrm{C}$ will be (A) \(5 \Omega\) (B) \(4 \Omega\) (C) \(3 \Omega\) (D) \(2 \Omega\)

Area of cross-section of a copper wire is equal to area of a square of $2 \mathrm{~mm}\( length It carries a current of \)8 \mathrm{~A}$ Find drift velocity of electrons (Density of free electrons in copper $\left.=8 \times 10^{28} \mathrm{~m}^{-3}\right)$ (A) \(1.56 \times 10^{-2} \mathrm{~ms}^{-1}\) (B) \(1.56 \times 10^{-4} \mathrm{~ms}^{-1}\) (C) \(3.12 \times 10^{-2} \mathrm{~ms}^{-1}\) (D) \(3.12 \times 10^{-3} \mathrm{~ms}^{-1}\)

An electrical circuit is shown in fig the values of resistances and the directions of the currents are shown A voltmeter of resistance \(400 \Omega\) is connected across the \(400 \Omega\) resistor. The battery has negligible internal resistance.The residing of the voltmeter is. \((\mathrm{A})(10 / 3) \mathrm{V}\) (B) \(5 \mathrm{~V}\) (C) \((20 / 3) \mathrm{V}\) (D) \(4 \mathrm{~V}\)

when a cell is connected to a resistance \(R_{1}\) the rate at which heat is generated in it is the same as when the cell is connected to a resistance \(\mathrm{R}_{2}\left(<\mathrm{R}_{1}\right)\) the internal resistance of the cell is.... (A) \(\left(R_{1}-R_{2}\right)\) (B) \((1 / 2)\left(\mathrm{R}_{1}-\mathrm{R}_{2}\right)\) (C) $\left\\{\left(\mathrm{R}_{1} \mathrm{R}_{2}\right) /\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)\right.$ (D) \(\sqrt{R}_{1} R_{2}\)

How would you arrange 48 cells each of e.m.f \(2 \mathrm{~V}\) and internal resistance \(1.5 \Omega\) so as to pass maximum current through the external resistance of \(2 \Omega\) ? (A) 2 cells in 24 groups (B) 4 cells in 12 groups (C) 8 cells in 6 groups (D) 3 cells in 16 groups
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