Question

A potential difference of 6 V is applied across the two terminals of an electric

bulb . If the resistance of the electric bulb is 10 Ω then calculate the strength of

electric current .
Pls answer then surely I will mark you brainleist. ​

Answer 1

Answer:

P.D. = 6V

R = 10 Ω

By ohm's law,

V = IR

6V = I * 10 Ω

I = 6/10

I = 0.6 A

Answer : 0.6 ampere

Explanation:

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Answer 2

Answer:

$\huge\boxed{\fcolorbox{red}{pink}{Your's Answer}}$

V= 6v

R= 10Ω

I= ??

$</em></strong><strong><em>R</em></strong><strong><em> </em></strong><strong><em>= \frac{</em></strong><strong><em>V</em></strong><strong><em>}{</em></strong><strong><em>I</em></strong><strong><em>}</em></strong><strong><em>$

(BY OHM'S LAW)

$</em></strong><strong><em>I</em></strong><strong><em> </em></strong><strong><em>= \frac{v}{</em></strong><strong><em>R</em></strong><strong><em>}$

$</em></strong><strong><em>I</em></strong><strong><em> = \frac{6}{10}$

$</em></strong><strong><em>I</em></strong><strong><em> </em></strong><strong><em>= \frac{3}{5}$

$</em></strong><strong><em>I</em></strong><strong><em> = 0.6</em></strong><strong><em>A</em></strong><strong><em>$

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