Physics A: Problem Set 9: Electric Circuits
recommended reading
Barron's Let's Review:  9.7 Series Circuits, 9.8 Parallel Circuits 
physics.info:  Resistors in Circuits 
Wikipedia:  Series and parallel circuits 
HyperPhysics:  Resistance 
Khan Academy:  Circuits (part 2), Circuits (part 3), Circuits (part 4) 
in class
 Determine the following quantities for the given circuit…
 the equivalent resistance,
 the total current from the power supply,
 the current through each resistor,
 the voltage drop across each resistor, and
 the power dissipated in each resistor.
Follow the rules for series circuits.
Resistances in series add up.
R_{T} = R_{1} + R_{2} + R_{3} R_{T} = 20 Ω + 30 Ω + 50 Ω R_{T} = 100 Ω Total current is determined by the voltage of the power supply and the equivalent resistance of the circuit.
I_{T} = V_{T}/R_{T}
I_{T} = 125 V/100 Ω
I_{T} = 1.25 ACurrent is constant through resistors in series.
I_{T} = I_{1} = I_{2} = I_{3} = 1.25 A
The voltage drops can be found using Ohm's law.
V_{1} = I_{1}R_{1}
V_{1} = (1.25 A)(20 Ω)
V_{1} = 25.0 VV_{2} = I_{2}R_{2}
V_{2} = (1.25 A)(30 Ω)
V_{2} = 37.5 VV_{3} = I_{3}R_{3}
V_{3} = (1.25 A)(50 Ω)
V_{3} = 62.5 VVerify your calculations by adding the voltage drops. On a series circuit they should equal the voltage increase of the power supply.
V_{T} = V_{1} + V_{2} + V_{3} 125 V = 25.0 V + 37.5 V + 62.5 V 125 V = 125 V We're good, so let's finish.
There are three equations for determining power. Since we have three resistors, let's apply a different equation to each as an exercise.
P_{1} = V_{1} I_{1}
P_{1} = (25.0 V)(1.25 A)
P_{1} = 31.250 WP_{2} = I_{2}^{2}R_{2}
P_{2} = (1.25 A)^{2}(30 Ω)
P_{2} = 46.875 WP_{3} = V_{3}^{2}/R_{3}
P_{3} = (62.5 V)^{2}/(50 Ω)
P_{3} = 78.125 WIn a series circuit, the element with the greatest resistance consumes the most power.
 Determine the following quantities for the given circuit…
 the equivalent resistance,
 the total current from the power supply,
 the current through each resistor,
 the voltage drop across each resistor, and
 the power dissipated in each resistor.
Follow the rules for parallel circuits.
Resistances in parallel combine according to the sumofinverses rule.
1 = 1 + 1 + 1 R_{T} R_{1} R_{2} R_{3} 1 = 1 + 1 + 1 R_{T} 20 Ω 100 Ω 50 Ω 1 = 5 + 1 + 2 R_{T} 100 Ω 100 Ω 100 Ω 1 = 8 R_{T} 100 Ω R_{T} = 100 Ω = 12.5 Ω 8 Total current is determined by the voltage of the power supply and the equivalent resistance of the circuit.
I_{T} = V_{T}/R_{T}
I_{T} = 125 V/12.5 Ω
I_{T} = 10 A
(Note: we'll answer part iv before part iii.) On a parallel circuit, each branch experiences the same voltage drop.
V_{T} = V_{1} = V_{2} = V_{3} = 125 V

The current in each branch can be found using Ohm's law.
I_{1} = V_{1}/R_{1}
I_{1} = (125 V)/(20 Ω)
I_{1} = 6.25 AI_{2} = V_{2}/R_{2}
I_{2} = (125 V)/(100 Ω)
I_{2} = 1.25 AI_{3} = V_{3}/R_{3}
I_{3} = (125 V)/(50 Ω)
I_{3} = 2.50 AVerify your calculations by adding the currents. On a parallel circuit they should add up to the current from the power supply.
I_{T} = I_{1} + I_{2} + I_{3} 10 A = 6.25 A + 1.25 A + 2.50 A 10 A = 10 A Good, it works.

Again as an exercise, use a different equation to determine the electric power of each resistor.
P_{1} = V_{1}I_{1}
P_{1} = (125 V)(6.25 A)
P_{1} = 781.25 WP_{2} = I_{2}^{2}R_{2}
P_{2} = (1.25 A)^{2}(100 Ω)
P_{2} = 156.25 WP_{3} = V_{3}^{2}/R_{3}
P_{3} = (125 V)^{2}/(50 Ω)
P_{3} = 312.50 WIn a parallel circuit, the element with the least resistance consumes the most power.
 A power source and four identical, numbered light bulbs are connected together as shown in the circuit on the right.
 Rank the light bulbs in order of brightness.
 Describe the effect that removing any one of the light bulbs has (if any) on the brightness of the other three.
 remove bulb 1
 remove bulb 2
 remove bulb 3
 remove bulb 4
R_{1} R_{2} R_{3} R_{4} a. b. i. × b. ii. × b. iii. × b. iv. × Just the answers. No explanation.
R_{1} R_{2} R_{3} R_{4} a. brightest dimmest dimmest brightest b. i. × off off off b. ii. dimmer × brighter dimmer b. iii. dimmer brighter × dimmer b. iv. off off off ×
at home
 State two reasons why household outlets are not wired together in series.
There are actually three reasons why you shouldn't wire outlets in your home in series (at least three): inconsistent current, inconsistent voltage, and the risk that a single point of failure could shut down an entire house.
When elements are added to a series circuit, the resistance goes up and the current goes down. When they are removed, the opposite happens — the resistance goes down and the current goes up. Electrical devices are designed to operate at a specific current (or a reasonable range of currents). If every time you added an element to a circuit in your home the current in every device changed, no device would be able to operate in a consistent manner. Turn on the toaster and the lights get dim. Turn off the lights and the TV overloads. It would be frustrating for you and bad for your devices.
Each element of a series circuit gets a fraction of the available voltage. Every element added to a series circuit divides this voltage into smaller and smaller pieces. One device on a 120 V line gets 120 V. Two devices have to share this voltage. If they're identical, they share the voltage evenly. If one has a higher resistance than the other, it gets the majority of the 120 V. If one device has a variable resistance and the other doesn't, the constant device it at the mercy of the variable device. It'll get whatever voltage is leftover. This arrangement makes no sense.
Any break in any one element in series circuit affects all the other elements. Blow one light bulb in your home and every device turns off. Even more frustrating, the lights go out and now you have to figure out which device in your house caused it.
 What is meant by the term "short circuit"? What makes a short circuit dangerous?
A short circuit is an unintended conducting path in an otherwise properly functioning circuit that allows current to bypass a path of reasonably high resistance and follow a path of exceptionally low resistance. An example of a short circuit in the home would be sticking a metal fork into an empty wall outlet or pouring soup into a television set. This new, unexpected path of low resistance permits the flow of a larger than usual amount of current through the device (and through the wires in the walls bringing current to the device). Large currents mean lots of electrical energy. Lots of electrical energy typically means lots of heat. Heat up your forkoutlet, or your soupTV, or the electrical lines in your walls and watch them catch fire. Watch your whole house catch fire. In this case, I agree with Dr. Frankenstein's monster, "Fire bad".
 What happens to the total current as resistors are added to a circuit…
 in series?
 in parallel?
This problem doesn't ask for any explanations, but I'm going to add them anyway.
Current decreases as resistors are added to a series circuit. Adding elements in series adds more and more obstacles that the charges need to avoid (or smash into and then go around). All this extra resistance means less current.
Current increases as resistors are added to a parallel circuit. Adding elements to a parallel circuit gives the moving charges more space to move around in. More freedom means less resistance and more current.
 A kitchen in North America has three appliances connected to a 120 V circuit with a 15 A circuit breaker: an 850 W coffee maker, a 1200 W microwave oven, and a 900 W toaster.
 Draw a schematic diagram of this circuit.
 Which of these appliances can be operated simultaneously without tripping the circuit breaker?
Outlets are wired in parallel so that the appliances on a circuit are independent of one another. Turning the coffee maker off will not result in the toaster turning off (assuming both were on at the same time). Each appliance will also get the same regulated voltage, which simplifies the design of electrical devices. The downside to this scheme is that the parallel currents can add up to dangerously high levels. A circuit breaker in series before the parallel branches can prevent overloads by automatically opening the circuit.
A 15 A circuit operating at 120 V consumes 1,800 W of total power.
P = VI = (120 V)(15 A) = 1,800 W
Total power in a parallel circuit is the sum of the power consumed on the individual branches.
coffee maker + microwave oven 850 W + 1,200 W 2,050 W microwave oven + toaster 1,200 W + 900 W 2,100 W toaster + coffee maker 900 W + 850 W 1,750 W On this circuit, only the coffee maker and toaster can be operated simultaneously. All other combinations will trigger the circuit breaker to open.
 Track workers in the New York City subway system don't get their electricity from wall outlets. They use the third rail. Since the voltage of the third rail is different from the voltage of a wall outlet in an American home, they often use modified or specialized electrical devices. For example, the temporary lights the workers use are mounted on strips with five bulbs — never four, never six, always five. This design choice is intentional.
Source: MTA New York City Transit.
 What is the voltage of a typical household outlet in the United States?
 What is the voltage of the third rail in the New York City subway system?
 Why do the work lights used in the New York City subway come is strips with five light bulbs? Be sure to state the basic wiring pattern (series or parallel) in your answer.
It depends on your source of information, but nominal voltages for US and Canadian households are around 120 V to 125 V AC. When I test the lines at work in Brooklyn, NY I typically get 124 V. I think that the National Electrical Code standard is 120 V (±6 V), but I can't find any official document to back this up.
The third rail in the New York City subway system is either 600 V or 625 V DC higher than ground. Again, it depends on the source you consult. I've never tested the third rail myself. That is certainly dangerous and illegal.
Five light bulbs wired in series would divide up the third rail voltage so that each bulb got the nominal residential voltage in the US. That's why five bulbs are used. For example, 600 V/5 = 120 V or 625 V/5 = 125 V. Either pairing of values seems pretty reasonable. This way residential light bulbs can be connected to the nonresidential voltage of the third rail without blowing out.