a cell phone battery has 9.00x10^-3kwh of energy stored in it. If the battery runs at 3.20V and draws an average of 0.375A of current, how long will the battery last?
Show your work, include formulas, substitution, and express your answer rounded correctly with appropriate units
6 years ago
Answered By Leonardo F
This is a classic question about electricity. We basically have to use two formulas:
1) $P=i.V$P=i.V
2) $E=P.t$E=P.t
Thr first one tells us that the product between electric current and voltage is equal to power. The second one tells us that the product between power and time equals to energy. Combining the two of them, we can do this:
$P=\frac{E}{t}$P=Et and substitute this result in the other equation:
$\frac{E}{t}=i.V$Et=i.V
Since we have the energy (E), the electric current (i) and the voltage (V), we can solve for the time (t):
Converting the energy to SI units:
$E=9\times10^{-3}kWh\times\left(\frac{3.6\times10^6J}{1kWh}\right)=32400J$E=9×10−3kWh×(3.6×106J1kWh)=32400J . Now, we have:
6 years ago
Answered By Leonardo F
This is a classic question about electricity. We basically have to use two formulas:
1) $P=i.V$P=i.V
2) $E=P.t$E=P.t
Thr first one tells us that the product between electric current and voltage is equal to power. The second one tells us that the product between power and time equals to energy. Combining the two of them, we can do this:
$P=\frac{E}{t}$P=Et and substitute this result in the other equation:
$\frac{E}{t}=i.V$Et =i.V
Since we have the energy (E), the electric current (i) and the voltage (V), we can solve for the time (t):
Converting the energy to SI units:
$E=9\times10^{-3}kWh\times\left(\frac{3.6\times10^6J}{1kWh}\right)=32400J$E=9×10−3kWh×(3.6×106J1kWh )=32400J . Now, we have:
$\frac{32400J}{t}=\left(0.375A\right)\times\left(3.20V\right)$32400Jt =(0.375A)×(3.20V)
$t=\frac{32400J}{\left(0.375A\right)\times\left(3.20V\right)}=27000s$t=32400J(0.375A)×(3.20V) =27000s
Converting the time to hours and showing three significant digits, we have:
$t=27000s\times\left(\frac{1h}{3600s}\right)=7.50h$t=27000s×(1h3600s )=7.50h
The battery lasts 7 hours and a half.