When an emergency transmitter uses 325 watts and a receiver uses 50 watts, how many hours can a 12.6 volt, 55 ampere-hour battery supply full power to both units?
1.8 hours.
\[\frac{\text{mAh}\times \text{v}}{1000}=\text{Wh}\]
\[\frac{55000 \times 12.6}{1000}=693\]
\[\frac{693}{375}=1.848\] \[\approx1.8\]
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What current will flow in a 6 volt storage battery with an internal resistance of 0.01 ohms, when a 3-watt, 6-volt lamp is connected?
0.4995 amps.
To find current, we will use Ohms law and the Power law. In each form, they are:
\begin{align} \text{Ohm's law: } E &=I \times R\\ I &= \frac{E}{R}\\ R &= \frac{E}{I}\\ \\ \text{Power law: } P &= E \times I\\ E &= \frac{P}{I}\\ I &= \frac{P}{E}\\ \end{align}
Remember P is power in watts, E is voltage in volts (electromotive force, technically), I is current in amps, and R is resistance in Ohms (\(\Omega\)).
We need to know the full resistance, so we need to solve first for R of the lamp; we'll combine Ohm's law and the power law to get that:
\begin{align} R &= \frac{E}{I}\\ I &= \frac{P}{E}\\ \\ R &= \frac{E}{\frac{P}{E}}\\ R &= E \times \frac{E}{P}\\ &= 6V \times \frac{6\text{V}}{3\text{W}}\\ &= 12 \Omega \end{align}
Now add the Internal Resistance 0.01 \(\Omega\)
\[ 12 + .01 = 12.01 \Omega \]
Back to Ohm's law:
\begin{align} I &= \frac{E}{R}\\ &= \frac{6\text{V}}{12.01\Omega}\\ &= 0.4995 \text{ Amps} \end{align}
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A ship RADAR unit uses 315 watts and a radio uses 50 watts.
If the equipment is connected to a 50 ampere-hour battery rated at 12.6 volts, how long will the battery last?
1 hour 43 minutes.
From: wp2ahg
Total wattage is 315 + 50 = 365 watts.
Voltage is 12.6 volts.
The battery is rated at 50 amp-hours.
50 amp-hours x 12.6 volts = 630 watts
630 watts / 365 watts = 1.76 or 1hr 43min
Calculate amperage using Ohm's Law:
I = P / E
I = 365 / 12.6 = 28.97 amps
Since the battery is rated at 50 amp-hours, it would supply power for (50 Ah / 28.97 A) or 1.72 hours, or 1 hour 43 minutes.
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If a marine radiotelephone receiver uses 75 watts of power and a transmitter uses 325 watts, how long can they both operate before discharging a 50 ampere-hour 12 volt battery?
1 1/2 hours
\(P = 400W = 75W + 325W\)
\(E = 12V\)
\begin{align}
\text{Ohm's law: }\\
I &= \frac{P}{E}\\
Amps &= \frac{Watts}{Volts}\\
\\
\end{align}
\[\frac{400\text{Watts}}{12\text{ Volts}} = 33.33\text{ Amps}\]
\[\frac{50\text{Ah}}{33.33\text{ amps}} = 1.5\text{ hours}\]
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A 6 volt battery with 1.2 ohms internal resistance is connected across two light bulbs in parallel whose resistance is 12 ohms each. What is the current flow?
0.83 amps.
Two 12 ohm lamps in parallel is 6 ohms of resistance.
Add another 1.2 ohms for battery internal resistance and you get 7.2 ohms total resistance.
Now, employing Ohm's Law,
\[\frac{6V}{7.2\text{ ohms}} = 0.83\text{ amps}\]
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A 12.6 volt, 8 ampere-hour battery is supplying power to a receiver that uses 50 watts and a RADAR system that uses 300 watts. How long will the battery last?
17 minutes or 0.3 hours.
From wp2ahg:
P = Total wattage is
\({50\text{ watts}} + 300{\text{ watts}} = 350{\text{ watts}}\)
E = Voltage is \(12.6{\text{ volts}}\)
Calculate amperage using Ohm's Law:
\[\text{I}=\frac{\text{P}}{\text{E}}\]
\[\text{I}=\frac{350\text{ watts}}{12.6\text{ volts}}={27.7\text{ amperes}}\]
An 8 Ah (amperes hours) battery with a 27.7 amperes load will last
\[\frac{8\text{ Ah}}{27.7\text{ A}}={0.288\text{ hours}}\]
\({0.288\text{ hours}} * {60\text{ min/hour}} = {17.28\text{ minutes}}\)
\[\frac{17.28\text{ minutes}}{60\text{ minutes per hour}}={0.35\text{ hours}}\]
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