# how much charge is transported.

• Jan 1st 2011, 06:13 PM
bigwave
how much charge is transported.
a constant current of $3A$ for $4$ hours is required to charge an automotive battery. If the terminal voltage is $10 + t/2\ V$, where $t$ in hours

(a) how much charge is transported as a result of the charging?

(b) how much energy is expended?

(c) the much does the charging cost? Assume electricity costs $9 cents/kWh$

first was wondering the why the expression $10 + t/2\ V$ is used

thanks again for help(Cool)
• Jan 1st 2011, 07:33 PM
Sudharaka
Quote:

Originally Posted by bigwave
a constant current of $3A$ for $4$ hours is required to charge an automotive battery. If the terminal voltage is $10 + t/2\ V$, where $t$ in hours

(a) how much charge is transported as a result of the charging?

(b) how much energy is expended?

(c) the much does the charging cost? Assume electricity costs $9 cents/kWh$

first was wondering the why the expression $10 + t/2\ V$ is used

thanks again for help(Cool)

Dear bigwave,

For a), use the definition of current. $I=\frac{dq}{dt}$ where I is the current, q is the charge and t is the time. In this case the current is constant. Therefore this gives, $q=It$

For, b) you should know that the energy expended is given by, $E=VIt$ where V is the terminal volatage. Hope you know have a idea why the terminal voltage expression is given.

For c), first find out the kilowatt hours that had been used because of the charging. The power is given by, $P=VI$. You will have to multiply this with the hours used. Then use the given cost per kWh to estimate the charging cost.

Hope you will be able to continue.
• Jan 2nd 2011, 10:18 AM
bigwave
Q=43.2kC
well for (a) using $\int_0^{14400}(3)t=43200$or $43.2k[C]$

which is the books ans. however don't see need for the intergal

but for (b) the books ans is $475K[J]$but $(10+\frac{14400}{2})(3)(14400)

=3.1147$
which is not the ans

actually don't know why the voltage expression $10+\frac{t}{2}[V]$ is used
• Jan 3rd 2011, 08:42 PM
SammyS
Quote:

Originally Posted by bigwave
well for (a) using $\int_0^{14400}(3)t=43200$or $43.2k[C]$

which is the books ans. however don't see need for the integral

Because $\displaystyle dq=I\cdot dt\ \to\ q=\int_0^{14400}(3)\,dt$

Quote:

Originally Posted by bigwave
but for (b) the books ans is $475K[J]$but $(10+\frac{14400}{2})(3)(14400)

=3.1147$
×10^8 which is not the ans

actually don't know why the voltage expression $10+\frac{t}{2}[V]$ is used

It's given because: $\displaystyle Power={{dE}\over{dt}}=I\cdot V$

This gives: $\displaystyle E=\int_0^{14400} I\cdot V \, dt = \int_0^{14400}(3)\cdot \left(10+{{t}\over{2}}\right) \, dt\,.$ Of course this will be in Joules.

For (c) use the fact that 1 Joule = 1 Watt·Second.

• Jan 4th 2011, 01:11 AM
henryjack
Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. For more details Search in Wikipedia.
• Jan 4th 2011, 02:39 AM
Ackbeet
Post # 5 here is totally irrelevant to this thread. Could a mod please delete?
• Jan 4th 2011, 09:30 AM
bigwave
where the "10+" came from
Quote:

Originally Posted by SammyS

This gives: $\displaystyle E=\int_0^{14400} I\cdot V \, dt = \int_0^{14400}(3)\cdot \left(10+{{t}\over{2}}\right) \, dt\,.$ Of course this will be in Joules.

I was curious where the "10+" came from in the voltage
• Jan 4th 2011, 04:06 PM
bigwave
475.2kJ
Quote:

Originally Posted by SammyS

This gives: $\displaystyle E=\int_0^{14400} I\cdot V \, dt = \int_0^{14400}(3)\cdot \left(10+{{t}\over{2}}\right) \, dt\,.$ Of course this will be in Joules.

this did not give me the answer of $475.2kJ$
which is the answer in the book...
• Jan 4th 2011, 04:32 PM
$\displaystyle\ 3\int_{0}^4{\left(10+\frac{t}{2}\right)}dt=3\left( 10[4]+\frac{4^2}{4}\right)=132\;\;Watt-hours$
$1Wh=3.6kJ$
$132(3.6)=475.2kJ$