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3Q

Expert-verifiedFound in: Page 793

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**You are to connect resistors R1**** and R2****, with ${\mathbf{R}}{\mathbf{1}}{\mathbf{>}}{\mathbf{R}}{\mathbf{2}}$****, to a battery, first individually, then in series, and then in parallel. Rank those arrangements according to the amount of current through the battery, greatest first.**

Parallel arrangement >$>\mathrm{R}2>\mathrm{R}1>$ Series arrangement.

$\mathrm{R}1>\mathrm{R}2$

**Use the equation of Ohm’s law. According to the Ohm’s law, ${\mathbf{I}}{\mathbf{=}}\frac{\mathbf{V}}{\mathbf{R}}$****, so if voltage is constant, then c****urrent through the battery is inversely proportional to resistance.**

Formulae are as follow:

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

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

For the parallel arrangement:

The equivalent resistance of parallel arrangement is less and given as follow:

$\frac{1}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{R}}_{1}}+\frac{1}{{\mathrm{R}}_{2}}\phantom{\rule{0ex}{0ex}}{\mathrm{R}}_{\mathrm{eq}}=\frac{{\mathrm{R}}_{1}+{\mathrm{R}}_{2}}{{\mathrm{R}}_{1}{\mathrm{R}}_{2}}$

So, I depends only on ${\mathrm{R}}_{\mathrm{eq}}$ and for parallel arrangement ${\mathrm{R}}_{\mathrm{eq}}$ must be less than ${\mathrm{R}}_{1}\mathrm{and}{\mathrm{R}}_{2}$. So,

The current through the parallel arrangement is greatest.

For the series arrangement:

The equivalent resistance of series arrangement is more and given as follow:

${\mathrm{R}}_{\mathrm{eq}}={\mathrm{R}}_{1}+{\mathrm{R}}_{2}$

So, I depends only on ${\mathrm{R}}_{\mathrm{eq}}$ and for series arrangement ${\mathrm{R}}_{\mathrm{eq}}$ must be more than ${\mathrm{R}}_{1}\mathrm{and}{\mathrm{R}}_{2}$.

So, the current through the series arrangement is the least.

For the individual arrangements

${\mathrm{R}}_{1}>{\mathrm{R}}_{2}$

As ${\mathrm{R}}_{1}$ is greater than ${\mathrm{R}}_{2}$. So, the current through ${\mathrm{R}}_{2}$ is greater than ${\mathrm{R}}_{1}$ because $\mathrm{I}\propto \frac{1}{\mathrm{R}}$

So, the rank will be as follow:

Parallel arrangement >$>{\mathrm{R}}_{2}>{\mathrm{R}}_{1}>$ Series arrangement.

Hence, the Parallel arrangement >$>{\mathrm{R}}_{2}>{\mathrm{R}}_{1}>$ Series arrangement.

Therefore, rank the current through the battery as it depends on total resistance. For parallel resistance, the equivalent resistance is less, so, the current is more. And for series arrangement, the equivalent resistance is more, so, the current is less.

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