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Chapter 12: Problem 1754

If \(\sigma_{1}, \sigma_{2}\), and \(\sigma_{3}\) are the conductance's of three conductor then equivalent conductance when they are joined in series, will be. (A) \(\sigma_{1}+\sigma_{2}+\sigma_{3}\) (B) $\left(1 / \sigma_{1}\right)+\left(1 / \sigma_{2}\right)+\left(1 / \sigma_{3}\right)$ (C) $\left\\{\left(\sigma_{1} \sigma_{2} \sigma_{3}\right) /\left(\sigma_{1}+\sigma_{2}+\sigma_{3}\right)\right\\}$ (D) None of these.

Short Answer

Expert verified
The equivalent conductance when they are joined in series is given by: $$ \sigma_{eq} = \frac{1}{\left( \frac{1}{\sigma_{1}} + \frac{1}{\sigma_{2}} + \frac{1}{\sigma_{3}} \right)} $$ Hence, the correct answer is (B).

Step by step solution

01

Understand the reciprocal property between resistance and conductance

Conductance (\(\sigma\)) is the reciprocal of resistance (R). Mathematically, this can be expressed as: $$ \sigma = \frac{1}{R} $$ And the resistance can be expressed as: $$ R = \frac{1}{\sigma} $$
02

Find the equivalent resistance for resistors connected in series

When resistors are connected in series, their equivalent resistance (R_eq) can be found by adding the individual resistances: $$ R_{eq} = R_{1} + R_{2} + R_{3} $$
03

Convert conductances to resistances

Using the reciprocal property, we can find the resistances corresponding to the given conductances: $$ R_{1} = \frac{1}{\sigma_{1}}, \quad R_{2} = \frac{1}{\sigma_{2}}, \quad R_{3} = \frac{1}{\sigma_{3}} $$
04

Plug in the expressions for the resistances into the series formula

Plug in the resistances found in step 3 into the formula for equivalent resistance for resistors connected in series: $$ R_{eq} = \frac{1}{\sigma_{1}} + \frac{1}{\sigma_{2}} + \frac{1}{\sigma_{3}} $$
05

Convert the equivalent resistance to equivalent conductance

Now, we will convert the equivalent resistance back to the equivalent conductance using the reciprocal property: $$ \sigma_{eq} = \frac{1}{R_{eq}} $$ Plugging in the expression for \(R_{eq}\) found in step 4: $$ \sigma_{eq} = \frac{1}{\left( \frac{1}{\sigma_{1}} + \frac{1}{\sigma_{2}} + \frac{1}{\sigma_{3}} \right)} $$
06

Choose the correct answer

The resulting expression for the equivalent conductance is: $$ \sigma_{eq} = \frac{1}{\left( \frac{1}{\sigma_{1}} + \frac{1}{\sigma_{2}} + \frac{1}{\sigma_{3}} \right)} $$ Comparing our result from step 5 with the given options: (A) \(\sigma_{1}+\sigma_{2}+\sigma_{3}\) (B) \(\left(1 / \sigma_{1}\right)+\left(1 / \sigma_{2}\right)+\left(1 /\sigma_{3}\right)\) (C) \(\left\\{\left(\sigma_{1} \sigma_{2} \sigma_{3}\right)/\left(\sigma_{1}+\sigma_{2}+\sigma_{3}\right)\right\\}\) (D) None of these. We can observe that our result corresponds to option (B), which is the correct answer.

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Most popular questions from this chapter

An infinite sequence of resistances is shown in the figure. The resultant resistance between \(\mathrm{A}\) and \(\mathrm{B}\) will be, when \(\mathrm{R}_{1}=1 \mathrm{ohm}\) and \(\mathrm{R}_{2}=2 \mathrm{ohm}\) (A) \(3 \Omega\) (B) \(2 \Omega\) (C) \(1 \Omega\) (D) \(1.5 \Omega\)

A potentiometer wire of length \(1 \mathrm{~m}\) and resistance \(10 \Omega\) is connected in series with a cell of e.m.f \(2 \mathrm{~V}\) with internal resistance \(1 \Omega\) and a resistance box of a resistance \(R\) if potential difference between ends of the wire is \(1 \mathrm{~V}\) the value of \(R\) is. (A) \(4.5 \Omega\) (B) \(9 \Omega\) (C) \(15 \Omega\) (D) \(20 \Omega\)

n identical cells each of e.m.f \(E\) and internal resistance \(r\) are connected in series An external resistance \(R\) is connected in series to this combination the current through \(\mathrm{R}\) is. (A) \(\\{(\mathrm{nE}) /(\mathrm{R}+\mathrm{nr})\\}\) (B) \(\\{(\mathrm{nE}) /(\mathrm{n} R+\mathrm{r})\\}\) (C) \(\\{\mathrm{E} /(\mathrm{R}+\mathrm{nr})\\}\) (D) \(\\{(\mathrm{nE}) /(\mathrm{R}+\mathrm{r})\\}\)

Two resistances \(\mathrm{R}_{1}\) and \(\mathrm{R}_{2}\) have effective resistance \(\mathrm{R}_{\mathrm{s}}\) when connected in sires combination and \(R_{p}\) when connected in parallel combination if $\mathrm{R}_{8} \mathrm{R}_{\mathrm{p}}=16\( and \)\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)=4\( the values of \)\mathrm{R}_{1}\( and \)\mathrm{R}_{2}$ are (A) \(2 \Omega\) and \(0.5 \Omega\) (B) \(1 \Omega\) and \(0.25 \Omega\) (C) \(8 \Omega\) and \(2 \Omega\) (D) \(4 \Omega\) and \(1 \Omega\)

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) \(2 \Omega\)
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