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Q10Q

Expert-verifiedFound in: Page 765

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Three wires, of the same diameter, are connected in turn between two points maintained at a constant potential difference. Their resistivity and lengths are p**** and L (wire A), **

The rank of the wires according to the rate at which the energy is transferred to the thermal energy within them, the greatest first is ${P}_{C}>{P}_{A}>{P}_{B}$

- Three wires are of the same diameter and connected to the same potential difference.
- Resistivity and length
*L*of wire A is p and*L* - Resistivity and length
*L*of wire B is 1.2p and 1.2*L* - Resistivity and length
*L*of wire C is 0.9p and*L*

**The rate of energy transfer or power is equal to the product of current and voltage. **

** **

**We find the rank of the resistance of the wires using the formula of resistance. Then, using the relation between the resistance and the rate at which energy is transferred we can rank the wires according to the rate at which energy is transferred to the thermal energy within them, the greatest first.**

Formulae:

The resistance value of a material, R = p*L/A* …(i)

Here, is resistance, is resistivity, *L* is the length of the conductor, *A *is the area of the cross-section of the conductor.

The rate of energy transferred per second, $P={V}^{2}/R$ …(ii)

is energy transfer rate or power, *V* is potential difference, *R* is resistance.

From equation (i), we can get that the resistance value as follows:

$R\alpha pL$ …(iii)

(Since,the same diameter of wires leads to same area.)

For wire A, the resistance can be given using equation (iii) for same area value as follows:

${R}_{A}=\frac{pL}{A}$

For wire B, the resistance can be given using equation (iii) for same area value as follows:

${R}_{B}=\frac{(1.2p)(1.2L)}{A}\phantom{\rule{0ex}{0ex}}=1.44\frac{pL}{A}$

For wire C, the resistance can be given using equation (a) for same area value as follows:

${R}_{C}=\frac{90.9)pL}{A}\phantom{\rule{0ex}{0ex}}=0.9\frac{pL}{A}$

So, the resistance relation is ${R}_{B}>{R}_{A}>{R}_{C}$

But, from equation (ii), the rate of energy transfer per second can be given as:

$E\alpha \frac{1}{R}$

So, the energy transfer rate relation is ${P}_{C}>{P}_{A}>{P}_{B}$

Therefore, the rank of the wires according to the rate at which energy is transferred to the thermal energy within them, the greatest first is ${P}_{C}>{P}_{A}>{P}_{B}$

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