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Q3CQ

Expert-verifiedFound in: Page 771

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**There is a voltage across an open switch, such as in Figure 21.43. Why, then, is the power dissipated by the open switch small.**

The power dissipated is small because the current is very small.

**The term "circuit" refers to a set of electrical connections.**

**Circuits are closed-loop or route systems that consist of a network of electrical components through which electrons can travel.**

Resistance of the switch when closed ${R}_{closed}=0\Omega $

Resistance of switch when open ${R}_{open}=\infty \Omega $

Emf of the battery is $E$

The internal resistance of the battery is $r$

External resistance connected in series with the battery is $\mathrm{R}$

Current in the circuit when the switch is closed ${\mathrm{l}}_{\mathrm{closed}}$

Current in the circuit when the switch is open role="math" localid="1655897485259" ${\mathrm{l}}_{\mathrm{open}}$

When the switch is open,

Equivalent resistance in series of the circuit given,

${R}_{eq}=R+r+{R}_{open}$

Using Ohm's law, the current in the circuit is given as,

${\mathrm{l}}_{\mathrm{open}}=\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{eq}}}\phantom{\rule{0ex}{0ex}}{\mathrm{l}}_{\mathrm{open}}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}+{\mathrm{R}}_{\mathrm{open}}}$

${}_{\mathrm{open}}=\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}+\yen}\phantom{\rule{0ex}{0ex}}{\mathrm{l}}_{\mathrm{open}}=\frac{\mathrm{E}}{\yen}\phantom{\rule{0ex}{0ex}}{\mathrm{l}}_{\mathrm{open}}=0\mathrm{A}$

Therefore, when the circuit is open, there is no current in the circuit.

The power dissipated is given as,

$\mathrm{P}={\left({\mathrm{l}}_{\mathrm{open}}\right)}^{2}{\mathrm{R}}_{\mathrm{eq}}$

Substituting the values in the above expression, we get,

$\mathrm{P}={\left(0\mathrm{A}\right)}^{2}{\mathrm{R}}_{\mathrm{eq}}\phantom{\rule{0ex}{0ex}}=0\mathrm{W}$

Hence, the power dissipated across the switch when the open comes out to be $0\mathrm{W}$.

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