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15P

Expert-verifiedFound in: Page 795

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**The current in a single-loop circuit with one resistance R is 5.0 A. When an additional resistance of 2.0 ${\mathbf{\Omega}}$**

The value of R is.

$\mathrm{I}=5.0\mathrm{A}\phantom{\rule{0ex}{0ex}}{\mathrm{R}}^{\text{'}}=2.0\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}{\mathrm{I}}^{\text{'}}=4.0\mathrm{A}$

**Write for emf of the battery when the circuit has only resistance R and when R’ is inserted in the circuit using Ohm’s law. Equating them will give the value of R.**

**Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.**

Formulae are as follow:

$\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}$

Where, *I* is current, *V* is voltage, *R* is resistance.

Let the battery be emf V. Then, Ohm’s law gives,

$\mathrm{V}=\mathrm{IR}\phantom{\rule{0ex}{0ex}}\mathrm{V}=5\mathrm{R}$

When R’ is inserted in the circuit in series with R, the total resistance of the circuit becomes R+R’.

Then emf is given by,

$\mathrm{V}={\mathrm{I}}^{\text{'}}\left(\mathrm{R}+{\mathrm{R}}^{\text{'}}\right)\phantom{\rule{0ex}{0ex}}\mathrm{V}=4\left(\mathrm{R}+2\right)$

Equating the equations 1) and 2) gives,

$5\mathrm{R}=4\mathrm{R}+8\phantom{\rule{0ex}{0ex}}\mathrm{R}=8.0\mathrm{\Omega}$

Hence, the value of R is $8.0\mathrm{\Omega}$.

Therefore, using Ohm’s law, the unknown resistance in the circuit can be found.

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