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R tot = R 1 + R p . size 12{R rSub { size 8{"tot"} } =R rSub { size 8{1} } +R rSub { size 8{p} } } {}

First, we find R p size 12{R rSub { size 8{p} } } {} using the equation for resistors in parallel and entering known values:

1 R p = 1 R 2 + 1 R 3 = 1 6 . 00 Ω + 1 13 . 0 Ω = 0 . 2436 Ω . size 12{ { {1} over {R rSub { size 8{p} } } } = { {1} over {R rSub { size 8{2} } } } + { {1} over {R rSub { size 8{3} } } } = { {1} over {6 "." "00" %OMEGA } } + { {1} over {"13" "." 0 %OMEGA } } = { {0 "." "2436"} over { %OMEGA } } } {}

Inverting gives

R p = 1 0 . 2436 Ω = 4 . 11 Ω . size 12{R rSub { size 8{p} } = { {1} over {0 "." "2436"} } %OMEGA =4 "." "11" %OMEGA } {}

So the total resistance is

R tot = R 1 + R p = 1 . 00 Ω + 4 . 11 Ω = 5 . 11 Ω . size 12{R rSub { size 8{"tot"} } =R rSub { size 8{1} } +R rSub { size 8{p} } =1 "." "00" %OMEGA +4 "." "11 " %OMEGA =5 "." "11 " %OMEGA } {}

Discussion for (a)

The total resistance of this combination is intermediate between the pure series and pure parallel values ( 20.0 Ω and 0.804 Ω , respectively) found for the same resistors in the two previous examples.

Strategy and Solution for (b)

To find the IR size 12{ ital "IR"} {} drop in R 1 size 12{R rSub { size 8{1} } } {} , we note that the full current I size 12{I} {} flows through R 1 size 12{R rSub { size 8{1} } } {} . Thus its IR size 12{ ital "IR"} {} drop is

V 1 = IR 1 . size 12{V rSub { size 8{1} } = ital "IR" rSub { size 8{1} } } {}

We must find I size 12{I} {} before we can calculate V 1 size 12{V rSub { size 8{1} } } {} . The total current I size 12{I} {} is found using Ohm’s law for the circuit. That is,

I = V R tot = 12 . 0 V 5 . 11 Ω = 2 . 35 A . size 12{I= { {V} over {R rSub { size 8{"tot"} } } } = { {"12" "." 0" V"} over {5 "." "11 " %OMEGA } } =2 "." "35"" A"} {}

Entering this into the expression above, we get

V 1 = IR 1 = ( 2 . 35 A ) ( 1 . 00 Ω ) = 2 . 35 V . size 12{V rSub { size 8{1} } = ital "IR" rSub { size 8{1} } = \( 2 "." "35"" A" \) \( 1 "." "00" %OMEGA \) =2 "." "35"" V"} {}

Discussion for (b)

The voltage applied to R 2 size 12{R rSub { size 8{2} } } {} and R 3 size 12{R rSub { size 8{3} } } {} is less than the total voltage by an amount V 1 size 12{V rSub { size 8{1} } } {} . When wire resistance is large, it can significantly affect the operation of the devices represented by R 2 size 12{R rSub { size 8{2} } } {} and R 3 size 12{R rSub { size 8{3} } } {} .

Strategy and Solution for (c)

To find the current through R 2 size 12{R rSub { size 8{2} } } {} , we must first find the voltage applied to it. We call this voltage V p size 12{V rSub { size 8{p} } } {} , because it is applied to a parallel combination of resistors. The voltage applied to both R 2 size 12{R rSub { size 8{2} } } {} and R 3 size 12{R rSub { size 8{3} } } {} is reduced by the amount V 1 size 12{V rSub { size 8{1} } } {} , and so it is

V p = V V 1 = 12 . 0 V 2 . 35 V = 9 . 65 V . size 12{V rSub { size 8{p} } =V - V rSub { size 8{1} } ="12" "." 0" V" - 2 "." "35"" V"=9 "." "65"" V"} {}

Now the current I 2 size 12{I rSub { size 8{2} } } {} through resistance R 2 size 12{R rSub { size 8{2} } } {} is found using Ohm’s law:

I 2 = V p R 2 = 9 . 65 V 6 . 00 Ω = 1 . 61 A . size 12{I rSub { size 8{2} } = { {V rSub { size 8{p} } } over {R rSub { size 8{2} } } } = { {9 "." "65 V"} over {6 "." "00 " %OMEGA } } =1 "." "61"" A"} {}

Discussion for (c)

The current is less than the 2.00 A that flowed through R 2 size 12{R rSub { size 8{2} } } {} when it was connected in parallel to the battery in the previous parallel circuit example.

Strategy and Solution for (d)

The power dissipated by R 2 size 12{R rSub { size 8{2} } } {} is given by

P 2 = ( I 2 ) 2 R 2 = ( 1 . 61 A ) 2 ( 6 . 00 Ω ) = 15 . 5 W . size 12{P rSub { size 8{2} } = \( I rSub { size 8{2} } \) rSup { size 8{2} } R rSub { size 8{2} } = \( 1 "." "61"" A" \) rSup { size 8{2} } \( 6 "." "00" %OMEGA \) ="15" "." 5" W"} {}

Discussion for (d)

The power is less than the 24.0 W this resistor dissipated when connected in parallel to the 12.0-V source.

Applying the science practices: circuit construction kit (dc only)

Plan an experiment to analyze the effect on currents and potential differences due to rearrangement of resistors and variations in voltage sources. Your experimental investigation should include data collection for at least five different scenarios of rearranged resistors (i.e., several combinations of series and parallel) and three scenarios of different voltage sources.

Practical implications

One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively large, as in a worn (or a very long) extension cord, then this loss can be significant. If a large current is drawn, the IR size 12{ ital "IR"} {} drop in the wires can also be significant.

For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see the passenger compartment light dim when you start the engine of your car (although this may be due to resistance inside the battery itself).

What is happening in these high-current situations is illustrated in [link] . The device represented by R 3 size 12{R rSub { size 8{3} } } {} has a very low resistance, and so when it is switched on, a large current flows. This increased current causes a larger IR size 12{ ital "IR"} {} drop in the wires represented by R 1 size 12{R rSub { size 8{1} } } {} , reducing the voltage across the light bulb (which is R 2 size 12{R rSub { size 8{2} } } {} ), which then dims noticeably.

Questions & Answers

Suppose a speck of dust in an electrostatic precipitator has 1.0000×1012 protons in it and has a net charge of –5.00 nC (a very large charge for a small speck). How many electrons does it have?
Alexia Reply
how would I work this problem
Alexia
how can you have not an integer number of protons? If, on the other hand it supposed to be 1e12, then 1.6e-19C/proton • 1e12 protons=1.6e-7 C is the charge of the protons in the speck, so the difference between this and 5e-9C is made up by electrons
Igor
what is angular velocity
Obaapa Reply
Why does earth exert only a tiny downward pull?
Mya Reply
hello
Islam
Why is light bright?
Abraham Reply
what is radioactive element
Attah Reply
an 8.0 capacitor is connected by to the terminals of 60Hz whoes rms voltage is 150v. a.find the capacity reactance and rms to the circuit
Aisha Reply
thanks so much. i undersooth well
Valdes Reply
what is physics
Nwafor Reply
is the study of matter in relation to energy
Kintu
a submersible pump is dropped a borehole and hits the level of water at the bottom of the borehole 5 seconds later.determine the level of water in the borehole
Obrian Reply
what is power?
aron Reply
power P = Work done per second W/ t. It means the more power, the stronger machine
Sphere
e.g. heart Uses 2 W per beat.
Rohit
A spherica, concave shaving mirror has a radius of curvature of 32 cm .what is the magnification of a persons face. when it is 12cm to the left of the vertex of the mirror
Alona Reply
did you solve?
Shii
1.75cm
Ridwan
my name is Abu m.konnek I am a student of a electrical engineer and I want you to help me
Abu
the magnification k = f/(f-d) with focus f = R/2 =16 cm; d =12 cm k = 16/4 =4
Sphere
what do we call velocity
Kings
A weather vane is some sort of directional arrow parallel to the ground that may rotate freely in a horizontal plane. A typical weather vane has a large cross-sectional area perpendicular to the direction the arrow is pointing, like a “One Way” street sign. The purpose of the weather vane is to indicate the direction of the wind. As wind blows pa
Kavita Reply
hi
Godfred
what about the wind vane
Godfred
If a prism is fully imersed in water then the ray of light will normally dispersed or their is any difference?
Anurag Reply
the same behavior thru the prism out or in water bud abbot
Ju
If this will experimented with a hollow(vaccum) prism in water then what will be result ?
Anurag
What was the previous far point of a patient who had laser correction that reduced the power of her eye by 7.00 D, producing a normal distant vision power of 50.0 D for her?
Jaydie Reply
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Jaydie
29/20 ? maybes
Ju
In what ways does physics affect the society both positively or negatively
Princewill Reply
Practice Key Terms 9

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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