**Problem Set 18:** | 1 | 3 | 5 | 7 | 9 | 23
| 25 | 27 | 29 | 35 | 41 | 43 | 44
| 45 | Go
Up

* - by Nina Murthy, April 2002*

**1. A service station charges a battery using a current of
5.7A for 7.0h. How much charge passes through the battery? **

We are given the amps and time and need to find the charge. The perfect equation for this is:

DQ = IDT

where I = 5.7c/s and T can be converted to seconds (multiply by 60 and then 60 again), to equal 25,200s. Then we just plug into the above equation to solve…

DQ = (5.7c/s)*(25,200s)

*DQ=1.4 E 5c
*

**3. What is the current in amperes if 1,000
Na+ ions were to flow across a cell membrane in 6.5****ms?
The charge on the sodium is the same as on an electron, but positive. **

Here we will use a different version of the equation DQ = IDT used above, as follows:

I=DQ /DT

and plug in the values given as Q=(1000)*(1.6E-19) and T=6.5E-6s, as follows:

I=( (1000)*(1.6E-19) ) / (6.5E-6s)

and then* I= 2.46E-11amps*

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**5. What is the resistance of a toaster if 110V produces a
current of 3.1A? **

Here we will use the equation R=V/I and just plug in as follows:

R=110V/3.1A

*R=35.48
ohms*

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**7. A 9.0V batter is connected to a bulb whose resistance
is 1.6ohms. How many electrons leave the battery per minute? **

Here we will use the same equation as in #5, only change it to be:

I=V/R

and plug in values as such: I=9/1.6 and calculate this to be 5.625.

We then incorporate the electrons by multiplying 5.625*(1/ (1.602E-19) ) and calculate this to equal 3.5E19.

The answer must be per minute
however, so we multiply this value by 60 and calculate *2.1E21electrons/min* to
be our answer.

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**9. If a 12V battery pushes a current of 0.50A through a
resistor, what is its resistance, and how many joules of energy does the batter
lose in a minute? **

The equation applied here is once again R=V/I and a straight substitution is done as follows:

*R=12V/0.5A
*

which equals 24ohms as the attracted resistance.

To calculate joules, the equation W=VIt is used and substituted for as follows:

W=12V*0.5amps*60s which is
calculated to be *360J*—the amount of
energy lost in one minute.

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**23. The element of an electric oven is designed to
produce 3.3kW of heat when connected to a 240-V source. What must be the
resistance of the element? **

The known values are as such:

P=3300W

V=240V

R= ?

Thus, an appropriate equation
seems to be P=V^{2}/R. We can then plug in the values we defined above:

3300=240^{2}/R

*R=17ohms*

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**25. What is the maximum voltage that can be applied to a
2.7k-ohm resistor rated at ¼ watt? **

Again we will define the known values:

V= ?

R=2700ohms

P=0.25W

Again, the equation P=V^{2}/R
will be used and substituted in for:

0.25=V^{2 }/2700

*V=26v*

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**27. a.) What is the resistance and current through a 60-W
light bulb if it is connected to its proper source voltage of 120V? **

Define variables as follows:

P=60W

V=120V

R= ?

and use the equation P=V^{2}/R.
Plug in values above as follows:

60=120^{2}/R

*R=240ohms*

To solve for current, use the equation P=IV where P=60 and V=120 and substitute as follows:

60=I*120

*I=0.5amps*

**b.) Repeat for a 440 W bulb.**

Define variables as follows:

P=440W

V=120V

R= ?

and use the equation P=V^{2}/R.
Plug in values above as follows:

440=120^{2}/R

*R=33ohms*

To solve for current, use the equation P=IV where P=440 and V=120 and substitute as follows:

440=I*120

*I=3.7amps
*

**29. How many Kwh of energy does a 550-W toaster use in
the morning if it is in operation for a total of 10min? **

Define known values:

P=550W

t=(10min)*(60s)=600s

Calculate as follows:

( (550)*(600) )/ (3.6E6J) which equals *0.092kWh*. (The
value of 3.6E6J is the value for one kWh.)

**--At a cost of 12cents/kWh, how much would this add to your monthly
electric energy bill if you made toast four mornings per week? **

(4mornings)*(4weeks)*(0.092kW)*(0.12cents)= *18cents*

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**35. What is the efficiency of a 0.50hp electric motor
that draws 4.4A from a 120-V line? **

P out=(0.5He)(745.7)=372.85watts

P in=(4.4)(120)=528watts

e= (P out)/(P in)

e=372.85watts/528watts

e=0.706 or *71% efficiency*

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**41. An ac voltage, whose peak value is 180V, is across a
330ohm resistor. What is the value of the rms and peak currents in the
resistor? **

Ohm's law is obeyed in a resistor
instantaneously. So if the peak voltage is 180 V, the peak current is
simply I = V/R, I_{o} = (180 V)/(330 ohms) = 0.545454545 = 0.545
amps.

For finding the rms current we just use the rms voltage.

so V_{rms} = V_{0}/Ö2

V_{rms} = 180/Ö2
= 127.3 V, and now we just use ohm's law:

I = V/R = (127.3 V)/(330 ohm) = 0.386 amps.

Notice that we also could have used

I_{rms} = I_{0}/Ö2

I_{rms} = (0.545454545)/Ö2=
0.386 A

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**43. The peak value of an alternating current passing
through a 1500W device is 4.0A. What is the rms voltage across it? **

To relate current and power using
the formula P = IV, the values of I and V must be rms values, so first we will
find the rms current:

I_{o} = 4.0A

I_{rms} = I_{o}/Ö2

I rms = (4.0A)/Ö2 which equals
2.828 A

The equation P=(V rms)*(I rms) is the applied and calculated as follows:

1500=(V rms)*(2.828) and therefore
*V rms=530v
*

**44. Calculate the peak voltage and peak current
through an 1800 W arc welder connected to a 450 V ac line.**

We are going to use the rms
equations to find the peak values:

I_{rms} = I_{0}/Ö2

V_{rms} = V_{0}/Ö2

We already have the rms voltage (450 V - remember, if it doesn't say peak, it
means rms), all we have to do is find the rms current, which is easy. P =
IV, so I = P/V:

Irms = (1800 W)/(450 V) = 4.0 A.

So now we plug them into the rms formulas:

I_{0}_{} = I_{rms}Ö2
= 5.657 A

V_{0}_{} = V_{rms}Ö2
= 636.4 V

It is interesting to note that the peak power (Which occurs at the simultaneous
peak of voltage and current) is __ twice__ the average power:

P

**45. What is the maximum instantaneous power
dissipated by, and maximum current passing through, a 3.0- HP pump connected to
a 240- V ac power source.**

First off, we need 3.0
horsepower in watts:

P = (3.0 HP)(745.699872 W/HP) = 2237.1 W

Now, if you looked at the last problem, you would know that the maximum or peak
power is simply equal to twice this, but here we will arrive at this conclusion
honestly. (using IB formulas)

We already know the rms voltage to be 240 V, so the peak voltage is:

V_{rms} = V_{0}/Ö2

V_{0} = (240 V)Ö2
= 339.41 V

We need the Irms, which we get from P = IV (I and V are by default and
necessarily rms to get the average power, P) so I = P/V = (2237.1 W)/(240 V) =
9.3212 A

And now we find the peak current:

I_{rms} = I_{0}/Ö2

I_{0} = (9.3212 A)Ö2
= 13.182 A

So the peak power is simply the peak current times the peak voltage (

P_{o} = I_{o}V_{o} = (13.182 A)(339.41 V) = 4474.2
W (This is exactly 6.0 hp, or twice the
average power. This happens because the square root of 2 squared is 2)

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