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Katyanochek1 [597]

3 years ago

10

Two identical light bulbs, each with resistance R = 2  are connected to a source with E = 8 V and negligible internal resistanc

e. Find the current through each bulb, if the bulb in series and (ii) in parallel.

1 answer:

Fiesta28 [93]3 years ago

8 0

R_1=R_2=2ohm
V=8V

<h3>In series</h3>

$\boxed{\sf R=R_1+R_2}$

$\\ \sf\longmapsto R=2+2$

$\\ \sf\longmapsto R=4\Omega$

<h3>In parallel</h3>

$\\ \sf\longmapsto \dfrac{1}{R}=\dfrac{1}{R_1}=+\dfrac{1}{R_2}$

$\\ \sf\longmapsto \dfrac{1}{R}=\dfrac{1}{2}+\dfrac{1}{2}$

$\\ \sf\longmapsto \dfrac{1}{R}=1$

$\\ \sf\longmapsto R=1\Omega$

Now

Using ohms law

$\boxed{\sf \dfrac{V}{I}=R}$

$\\ \sf\longmapsto I=\dfrac{V}{R}$

In series:-

$\\ \sf\longmapsto I=\dfrac{8}{4}$

$\\ \sf\longmapsto I=2A$

<h3>In parallel</h3>

$\\ \sf\longmapsto I=\dfrac{8}{1}$

$\\ \sf\longmapsto I=8A$

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Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

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