Billions In The Blink Of An Eye

From $\text{power} = {I}{V}$ $\text{current}=\frac{P}{V}$ $\text{current}={0.5A}$

We know that charge is quantized, that is, q is an integral multiple of e , thus $Q={I}{t}={q}{n} \longrightarrow t =\frac{qn}{I}$ $\text{Time} = \frac{({1.6}\times{10}^{-19})\times({1}\times{10}^9)}{0.5}$ Therefore $\text{time (in seconds)} =\boxed{3.2\times10^{-10}}$

Using usual notations,

$P=V*I$

Therefore, $I=\frac{60}{120}=0.5A$

Also,

$I=\frac{Q}{t}$

Now, charge of 1billion electrons = $10^{9}*1.6*10^{-19}$

Thus, $t=\frac{1.6*10^{-10}}{0.5}$

$=\boxed{3.2*10^{-10} sec}$

To start answering this question we need to understand what is meant by the unit 'Watt'. A watt is one joule per second - that, is the amount of energy used whilst trying to do something ('work done') per second.

It turns out that the unit 'Joule' is also a compound unit that can be broken down - it is equivalent to coulombs multiplied by volts. A coulomb is a unit of electrical charge, and a volt is basically the force that sends charges through a circuit. Now we know this, we can do some maths!

Let's express the above mathematically:

$W = \frac{J}{s}$ $J = C \times V$ $W = \frac{C \times V}{s}$

Now the problem looks a bit more manageable and seems to be a simple case of rearranging the formula to solve $s$ and plugging in values....but wait! The question asks about electrons, not these coulomb things! Fear not, because the charge of an electron can be expressed in coulombs. As correctly stated in the details and assumptions section, the charge of an electron is $e = 1.6 \times 10^{-19}$ Now we can rearrange the equation and answer the question!

$W = \frac{C \times V}{s}$ rearranges to: $s = \frac{C \times V}{W}$

There are 60W, 120V and one billion electrons (= $10^9 \times 1.6 \times 10^{-19} C$ )

$s = \frac{( (10^9 \times 1.6 \times 10^{-19})\times 120}{60}$

$s = \frac{1.92 \times 10^8}{60}$

$s = 3.2 \times 10^{-10}$

So the answer is $\boxed{3.2E-10}$ !

Let the power be $P = \text {60 W}$ . Let the voltage be $V = \text {120 V}$ . Then, the current is $I = \dfrac {P}{V} = 0.5 \, \text{A} = 0.5 \, \text{C/s}$ . Since we want the time it takes for $10^9$ electrons to flow, we convert to Coulombs. This is $1.6 \times 10^{-10} \, \text{C}$ . Thus, the amount of time is $t = \dfrac {Q}{I} = \dfrac {1.6 \times 10^{-10} \, \text{C}}{0.5 \, \text{A}} = \boxed {3.2 \times 10^{-10} \, \text{s}}.$

We have been given-

$W = 60$ $watts$ and $V = 120$ $volts$ .

We know that-

$q=ne$ ,

Where $q$ = Total Charge

$n$ = Number of electrons

$e$ = Charge of one electron

Therefore,

$q=10^9 \times 1.6 \times 10^{-19} = 1.6 \times 10^{-10}$

We know that, $W = VI$

Since $I = \frac{q}{t}$

We get-

$W = \frac{Vq}{t}$

Putting the values, we get-

$60 = \frac{120 \times 1.6 \times 10^{-10}}{t}$

Simplifying this, we get-

$t = 3.2\times 10^{-10} seconds.$

Therefore, the time taken is $3.2\times 10^{-10} seconds.$

Power (P)= 60 W

P.D. (V)=120 V

Number of electrons (n)= 1000000000

now by using, Power (P)=Potential difference (V) \times Current(I)

                 Current (I)= Power(P)/ (V)

                 Current (I)= 60/120

                               =1/2

                               =0.5 ampere

Now by using ....

Charge (Q) = Number of electrons \times charge on each electron

             = 10^{9} \times 1.6 \times 10^{-19}

             = 1.6 \times 10^{-10} coulomb

now by sing Current (I) = Charge (Q) / Time (t)

               Time (t) = Current (I) / Charge (Q)

                           = 0.5 / 1.6 \times 10^{-10}  

                           = 0.3125 \times 10^{10}

                           = 3125000000 sec

Power = Current x Voltage $P=I \times V$ But $I = \frac{Q}{t} = \frac{ne}{t}$

Therefore, $P = V\frac{ne}{t}$

Hence $t = ne\frac{V}{P}$

Plugging in the values for n = $10^9$ , e = $1.6 \times 10^{-19}$ C, V = 120 V and P = 60 W, we get $t = 3.2 \times 10^{-10}$