Problem Set 19 | 1 | 3 | 5 | 7 | 9 | 11 | Go Up

By Carpenter, 2004 Back to Physics homepage

**1. Four 140 W
light bulbs are connected in series. What is the total
resistance of the circuit? What is their resistance if they are
connected in parallel?**

-To start out with, light bulbs act as resistors so we will draw this diagram just like other circuits

-Next
we just add the resistance up to get the total resistance

140W + 140W +
140W + 140W =
**560 W**

**-**The next step is to calculate the parallel resistance, so
we re-construct the circuit

**-**Finally, we calculate the resistance with the formula for
resistance in a parallel circuit

1/(1/R + 1/R + 1/R + 1/R) = R’

1/(1/140W + 1/140W
+ 1/140W + 1/140W)
= **35W
TOP**

**3.** **Given only one 30W
and one 50W resistor, list all possible
values of resistance that can be obtained**

-so you can use the 30W by itself, and the 50W by itself

-you can also use both in a series (add em’ up)

-and you can use them in parallel (1/(1/30W+1/50W)

So the possible values are **30W,
50W, 80W, and
19W
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**5. Suppose that you have a 6.0 V
battery and you wish to apply a voltage of only 4.0 V. Given an
unlimited supply of 1.0W resistors, how
could you connect them so as to make a “voltage divider”
that produced a 4.0 V output from a 6.0 V input.**

Since series is the easiest to work with, lets try that first

So we want a circuit such that a segment of that circuit (multiple resistors) will produce four volts

Since two goes into both four and six, and when multiplied by one will yield voltage increments of 2, lets make 2 our current.

I=V/R = 2Amps = 6/ X (1W
resistors) ® X = 6/2 = **3 resistors
**
**TOP**

**7. What is the net resistance of the
circuit connected to the battery in the circuit below? Each
resistor has R= 2800W.**

This
is just a matter of simplifying the circuit, all resistors equal
2800W unless otherwise specified

-Series, so add 2800W + 2800W = 5600W

These two are in parallel so 1/(1/5600W + 1/2800W) = 1866.66666W

Series again so add em’, 1866.66666W + 2800W = 4666.66666W

More parallel, 1/ (1/4666.66666W + 1/2800W) = 1750W

Finally, another series so 1750W +
2800W = **4550W
or 4.6 kW
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**9. Eight lights are connected in
parallel to a 110 V source by two leads of total resistance 1.5W.
If .240 Amp current flows through each bulb, what is the resistance
of each, and what fraction of the total power is wasted in the leads?
**

Here is what we know**- **

First
we will calculate the current of this circuit

At the 1.5 W resistor (the lead) the current is equal to the sum of all the individual light bulb currents

Total current = .24A * 8 = 1.92A

Next we will find the voltage drop across the leads:

V = IR = 1.5W * 1.92A = 2.88V

We then subtract the amount of voltage that the leads drop from the source voltage to get the voltage across the light bulbs:

V _{total} – V _{leads} = V _{lights}
= 110V - 2.88V = 107.12V

So 107.12V is across each light bulb and now we can solve for individual bulb resistance

R = V/I = 107.12V/.24A = 446.33333 or **450W**

To calculate the power wasted at the leads, we will find the power each lead (.75*2W each) / total power

Power lost in leads = (1.92 A)^{2}(1.5W)
= 5.53 W

Total power consumed: P = IV = (1.92 A)(110 V) = 211.2 W

Percentage power lost = (5.53 W)/(211.2 W) = 0.026181818 or
**2.6%
****TOP**

**11. A close inspection of an
electric circuit reveals that a 480W
resistor was inadvertently soldered in place where a 320W
resistor is needed. How can this be fixed without removing
anything from the circuit.**

-so one of the child laborers in Taiwan was trying to get back at his evil master and put a resistor that had too much resistance into the circuit. Adding resistors in series will only make the resistance larger, so we must add circuits in parallel to equal 320W resistance.

-Since R_{parallel} = 1/(1/R+1/R), we fill in for what we
know

320W=1/(1/480W+1/R)

1=320W (1/480W+1/R)

1=320W/480W+320W/R

1=2W/3W+320W/R

1-(2W/3W)=320W/R

(1-(2W/3W))R=320W

R=320W/(1-(2W/3W))

**R=960
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