Question

 A current of 2.7A is drawn by a filament of an electric bulb for 15 minutes. Find the amount of electric charge that flows through the circuit. *

2 points


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Answer 1

$\large{\bf{\gray{\underline{\underline{\orange{Given:}}}}}}$

A current of 2.7A is drawn by a filament of an electric bulb for 15min.

$\large{\bf{\gray{\underline{\underline{\green{To\:Find:}}}}}}$

We have to find the amount of electric charge that flows through the circuit in the given interval of time.

$\large{\bf{\gray{\underline{\underline{\pink{Solution:}}}}}}$

◈ What is electric current !

Electric current is defined as the rate of charge flow per unit time.

SI unit : A or C/s

Mathematically,

$\dag\:\:\boxed{\bf{\pink{I=\dfrac{Q}{t}}}}$

❅ Conversion :

15min = 15 × 60 = 900s

Let's calculate ☃

⇒ I = Q/t

⇒ 2.7 = Q/900

⇒ Q = 2.7 × 900

⇒ Q = 2430C

Answer 2

Answer:

The amount of electric charge that flows through the circuit is 2430 C.

Step by step explanation-

Given-

• I ( current ) = 2.7A

• t ( time ) = 15 minutes = 900 seconds

Converting time to its SI unit ( seconds )

$\implies$ 15 × 60 = 900 seconds

To find-

• Q ( charge ) = ?

Solution-

We know that,

${\boxed {\rm {\red { I = \frac{Q}{T} }}}}$

But here we have to find Q ( charge ) , so from this formula we can say that

${\huge {\boxed {\rm {\red { Q= T \times I }}}}}$

Substituting values-

$\rm\blue { Q = 900 s \times 2.7A }$

$\rm\blue { Q = 2430 C }$

_________

Additionally-

Few formulae their values and Si units...

$\huge {\boxed {\green { I = \frac{Q}{T} }}}$

$\huge {\boxed {\green { V = \frac{W}{Q} }}}$

$\huge {\boxed {\green { V = I R }}}$

[ According to Ohm's law ]

______..

Q $\longrightarrow$ charge $\longrightarrow$ Coloumb ( C )

t $\longrightarrow$ time $\longrightarrow$ seconds

I $\longrightarrow$ current $\longrightarrow$ ampere ( A )

V $\longrightarrow$ potential difference $\longrightarrow$ volt (V)

W $\longrightarrow$ work done $\longrightarrow$ Joule ( J )

R $\longrightarrow$ resistance $\longrightarrow$ Ohm ( Ω )

_______...