Parts of a Circuit
The part of a circuit known to limit the flow of the electrical current.
Resistor
In a series circuit, electricity has ______ path(s) to follow.
one
In a parallel circuit, electricity has ______ path(s) to follow
multiple
What does Kirchhoff's Voltage Rule state?
Kirchhoff's Voltage rule states that the sum of all voltages in a loop must equal 0.
Is it better to wire your home appliances in parallel or series?
Parallel
What enables charge to move around in a circuit? What part of a circuit is this from?
The voltage difference across a circuit from the voltage source "pushes" charges through the circuit.
Imagine a series circuit that is connected to a 9V battery and has two 20 ohm resistors in series with each other. What is the equivalent current for this circuit?
9/40 A or 0.225A
Imagine a circuit supplied with 7 V and consists of resistors a, b, and c, with resistance values of 4Ω, 7Ω, and 10Ω, respectively. The three resistors are wired to be in parallel with each other. What would be the voltage drop across each resistor, and what is the equivalent resistance of the circuit?
Voltage drop across each resistor is 7V since voltage is the same across all points in a parallel circuit.
To solve for equivalent resistance in a parallel circuit we add in reciprocals:
1/Req=1/4Ω+1/7Ω+1/10Ω=69/140Ω
Req=140/69Ω= 2.03Ω
Imagine a circuit in which a 5Ω resistor is in series with [a parallel circuit where two 10Ω resistors are in parallel with each other]. Given that the voltage source has a voltage of 10V. How much power is dissipated through the 5 Ohm resistor?
total R of circuit = 5Ω + 5Ω = 10Ω.
V=IR, Icircuit = V/R = 10V/10Ω = 1A.
P of 5Ω resistor = I squared times R = 1A squared * 5Ω = 5W.
Why do electricians use wire that is coated in plastic when they are making new circuits?
This is because the wire can be dangerous to handle and can shock humans if they handle it when the circuit is on. The plastic coating helps to protect the electricians and us!
Explain how a switch works in an electrical circuit.
When the switch is on the circuit is closed so electricity can flow. When the switch is off the circuit is open so electricty cannot flow.
A circuit is wired with one cell supplying 5 V in series with three resistors of 3 Ω, 5 Ω, and 7 Ω, resistors labelled a, b, c, respectively. What is the resulting voltage across and current through each resistor of this circuit, as well as the entire circuit?
The total resistance of the resistors is R= 3Ω + 5Ω + 7Ω = 15Ω.
Now use Ohm’s law to get the current through the entire circuit:
I = V/R = 5V/15Ω = 0.33 A.
Because everything is in series, this is also the current through each circuit element.
Now, use Ohm’s law for each of the resistors in turn. The voltage drop across Ra = IR = 0.33 A * 3Ω = 1.0 V.
The voltage drop across Rb = IR = 0.33A * 5Ω =1.67 V.
The voltage drop across Rc = IR = 0.33A * 7Ω = 2.33 V.
A circuit is supplied voltage with a battery and has three resistors a, b, and c in parallel with each other, with resistance values of 5Ω, 3Ω, 6Ω respectively. If the current through a is 2A, what is the current through the rest of the resistors? What is the voltage from the battery?
Resistor a -> I=2A R=5Ω V=IR=2*5=10V
The voltage through the battery is 10V since voltage is same everywhere in a parallel circuit.
Resistor b -> V=10V R=3Ω I=V/R=10/3 A
Resistor c-> V=10V R=6Ω I=V/R=10/6=5/3 A
If the resistance of a wire in household appliance becomes 4 times its original value, which of the following statements is/are correct?
I. The voltage of the wire becomes quadrupled.
II. The current through the wire becomes 1/4th the original value.
III. The power consumed by the appliance becomes quadrupled.
only II.
I is false. The voltage is the electromotive force that drives the current through the wire. Therefore, it remains constant and unaffected by the increase in resistance. According to the equation V = IR, when V is constant, the current I is inversely proportional to the resistance R.
III is false. The power P is expressed by the equation P = IV, where I is the current through the wire and V is the voltage that drives the current. Since V stays constant, and I becomes 1⁄4 of the original value, the power P would be 1⁄4 I * V = 1⁄4 IV, which is 1⁄4 the original power output. II is true. V=IR. Iold = V/R. Since V is unchanged and R has quadrupled, Inew =V/4R = 1/4*Iold.
What are the SI units of Current?
Coulombs/s
Imagine a circuit with a 10V voltage source where a 1/2Ω resistor is in series with [a parallel circuit where one branch has a 2Ω resistor and the other has a 2/3Ω resistor]. Find the voltage drop across the 2/3Ω resistor.
Total R of parallel circuit is 1/2Ω. Add the R of 1/2Ω Resistor, so Total R of circuit is 1Ω. V=IR, I = 10A. Across 1/2Ω Resistor in series with battery, V = IR = 0.5Ω*10A = 5V. 10V battery - 5V series resistor means 5 volts is dropped across parallel part of circuit. Since V is same in either branch of parallel circuit, the voltage drop across 2/3Ω resistor is 5V.
A circuit is supplied with 8V, and has 2 resistors, a and b, wired in series with each other. Resistor a has resistance of .7Ω, but the resistance of resistor b is unknown. If the total resistance of the circuit is 4A, what is the resistance of resistor b? What is the voltage drop across each resistor?
V=IR -> R=V/I Req=8/4A=2Ω
Req=0.7Ω+Rb
Rb=2Ω-0.7Ω
Rb=1.3Ω
Va=4*.7=2.8V
Vb=4*1.3=5.2V
A kitchen in North America has three appliances connected in parallel 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.Which of these appliances can be operated simultaneously without tripping the circuit breaker?
Total power in a parallel circuit is the sum of the power consumed on the individual branches.
coffee maker+microwave oven =850W+1200W=2050W
microwave oven+toaster=1200W+900W=2100W
toaster +coffee maker=900W+850W=1750 W
Power of circuit=120*15=1800W
so only toaster and coffee maker will work
Daily Double! Imagine a series circuit with components in series in the following order clockwise: Voltage source going - to + of 30V, 3Ω Resistor, Voltage source going - to + of 10V, 5Ω Resistor, Voltage source going + to - of 8V. What is the current in the circuit?
30V - 3I + 10V - 5I - 8V = 0
32V = 8Ω*I
I = 4A
What type of meter should be used to measure small amounts of current?
a. Oscilloscope b. Voltmeter c. Ohmmeter
d. Galvanometer e.Ammeter
d. Galvanometer
You have a parallel circuit with 2 branches. On one branch is an open switch and a 10 Ω resistor. On the other branch is a 5 Ω resistor. The circuit is being supplied 10V. How does current on the battery change after the open switch is closed? By how much does it change?
When switch is open
V=10 Req=5 Ω I=V/R=10/5=2A
When switch is closed
1/Req=1/10+1/5=3/10 Ω -> Req=10/3 Ω
V=10V I=V/R=10/10/3=10*3/10=3A
The current on the battery increases by 1 A after the switch is closed.
You have 2 circuits, each with 1 battery, 1 resistor and 1 lightbulb. The first circuit is supplied 5V, and has a 100Ω resistor wired in series with a lightbulb that has a resistance of 350Ω. The second circuit is supplied 7V, and has a 20Ω resistor wired in series with a lightbulb that has a resistance of 400Ω. Which lightbulb is brighter, how do you know, and by how much?
To find brightness, we have to find power, P=IV.
Start by finding the current in each lightbulb.
Req1=100Ω+350Ω=450Ω
I1=5/450=1/90A
Req2=20Ω+400Ω=420Ω
I2=7/420Ω=1/60A
P1=5V*1/90A=5/90W
P2=7V*1/60A=7/60W
Bulb 2 is brighter by 0.061W
Imagine a parallel circuit with 7 identical resistors: a, b, c, d, e, f, g, h. The voltage supplied to the circuit is 9V. The current on one resistor is 2A. What is the equivalent current? What is the resistance of each resistor? What is the equivalent resistance?
Equivalent current -> 7(2A)=14A
Resistance on each resistor: 9/2=4.5 Ω
1/Req=7(1/4.5)=7/4.5 Ω
Req=4.5/7 Ω=45/70 Ω=9/14 Ω=0.643 Ω
Imagine a junction. 15.82 A flow into the junction, and 12.96 A flow out. What is wrong with this statement. How much current needs to flow out the junction to make this statement true?
Same amount of current going is not coming out. Need to use Kirchhoff's Current Rule. 2.86 A need to flow out the junction to make it true.
What was Volta's first battery called and how was it created?
Voltaic pile, alternating disks of zinc and silver separated by paper or cloth soaked in either saltwater or sodium hydroxide