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# Chapter 14: Basic Physics SE-DC Electric Circuits

Difficulty Level: At Grade Created by: CK-12

The Big Ideas:

The name electric current is the name given to the phenomenon that occurs when an electric field moves down a wire at close to the speed of light. Voltage is the electrical energy density that causes the electric current. Resistance is the amount a device in the wire resists the flow of current by converting electrical energy into other forms of energy. A device, the resistor, could be a light bulb, transferring electrical energy into heat and light or an electric motor that converts electric energy into mechanical energy. The difference in Voltage across a resistor or other electrical device is called the voltage drop.

In electric circuits (closed loops of wire with resistors and voltage sources) energy must be conserved. It follows that the algebraic sum of voltage drops and voltage sources, around any closed loop, will equal zero.

In an electric junction there is more than one possible path for current to flow. For charge to be conserved at a junction the current into the junction must equal the current out of the junction.

Key Equations:

• \begin{align*}I = \frac{\Delta q}{\Delta t}\end{align*} - current is the rate at which charge passes by; the units of current are Amperes \begin{align*}(1 \ A = 1 \ C/s)\end{align*}.
• \begin{align*}\Delta V = I \cdot R\end{align*} - the current flow through a resistor depends on the applied electric potential difference across it; the units of resistance are Ohms \begin{align*}(1 \ \Omega = 1 \ V/A)\end{align*}.
• \begin{align*}P = I \cdot \Delta V\end{align*} - the power dissipated by a resistor is the product of the current through the resistor and the applied electric potential difference across it; the units of power are Watts \begin{align*}(1 \ W = 1 \ J/s)\end{align*}.
• \begin{align*}E = P \cdot t\end{align*} - the electrical energy used is equal to the power dissipated multiplied by the time the circuit is running

Key Concepts:

• Ohm’s Law \begin{align*}V=IR\end{align*} (Voltage drop equals current times resistance.) This is the main equation for electric circuits but it is often misused. In order to calculate the voltage drop across a light bulb use the formula: \begin{align*}V_{lightbulb} = I_{lightbulb}R_{lightbulb}\end{align*}. For the total current flowing out of the power source, you need the total resistance of the circuit and the total current: \begin{align*}V_{total} = I_{total}R_{total}\end{align*}.
• Power is the rate that energy is released. The units for power are watts (W), which equal joules per second \begin{align*}[W] = [J]/[s]\end{align*}. For example, a 60 W light bulb transforms 60 joules of electrical energy into light and heat energy every second.

The equation used to calculate the power dissipated in a circuit is \begin{align*}P=IV\end{align*}. As with Ohm’s Law, one must be careful not to mix apples with oranges. If you want the power of the entire circuit, then you multiply the total voltage of the power source by the total current coming out of the power source. If you want the power dissipated (i.e. released) by a light bulb, then you multiply the voltage drop across the light bulb by the current going through that light bulb.

Name Symbol Electrical Symbol Units Water Analogy Everyday device
Voltage \begin{align*}V\end{align*}

volts \begin{align*}(V)\end{align*}

\begin{align*}V=J/C\end{align*}

A water dam with pipes coming out at different heights. The lower the pipe along the dam wall, the larger the water pressure, thus the higher the voltage. Battery, the plugs in your house, etc.
Current \begin{align*}I\end{align*}

amps \begin{align*}(A)\end{align*}

\begin{align*}A = C/s\end{align*}

A river of water. Objects connected in series are all on the same river, thus receive the same current. Objects connected in parallel make the main river branch into smaller rivers. These guys all have different currents. Whatever you plug into your wall sockets draws current
Resistance \begin{align*}R\end{align*}

_____

ohms \begin{align*}(\Omega)\end{align*} If current is analogous to a river, then resistance is the amount of rocks in the river. The bigger the resistance, the lower the current. Light bulb, Toaster, etc.
• Resistors in Series: All resistors are connected end to end. There is only one river, so they all receive the same current. But since there is a voltage drop across each resistor, they may all have different voltages across them – the sum of the voltage drops will equal the total voltage of the circuit. The more resistors in series the more rocks in the river, so the less current that flows.
\begin{align*}R_{total} = R_1 + R_2 + R_3 + \ldots\end{align*}
• Resistors in Parallel: All resistors are connected together at both ends. There are many rivers (i.e. the main river branches off into many other rivers), so all resistors receive different amounts of current. But since they are all connected to the same point at both ends they all receive the same voltage.
\begin{align*}\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots\end{align*}
• Direct Current (DC): Voltage pushes (so current flows) in one direction. Examples are batteries and the power supplies we use in class. Any time there’s a boxy device you plug into the wall (like a charger), its purpose is to convert AC from the wall to DC.
• Alternating Current (AC): Voltage pushes (so current flows) in alternate directions, back and forth. In the US they reverse direction 60 times a second (60 Hz). AC is more convenient than DC for transporting electrical energy. Below is a plot of voltage vs. time for a standard circuit in the USA.

Note: your TV and computer screen are actually flickering at 60 Hz due to the AC supplied by household plugs. Our eyesight does not work this fast, so we never notice it. However, if you film a TV the effect is observable due to the mismatched frame rates of the camera (usually 24 fps) and screen.

• Ammeter: A device that measures electric current. You must break the circuit to measure the current. Ammeters have very low resistance; therefore you must wire them in series.
• Voltmeter: A device that measures voltage. In order to measure a voltage difference between two points, place the probes down on the wires for the two points. Do not break the circuit. Voltmeters have very high resistance; therefore you must wire them in parallel.
• Voltage source: A power source that produces fixed voltage regardless of what is hooked up to it. A battery is a real-life voltage source. A battery can be thought of as a perfect voltage source with a small resistor (called internal resistance) in series. The electric energy density produced by the chemistry of the battery is called emf (electromotive force), but the amount of voltage available from the battery is called terminal voltage. The terminal voltage equals the emf minus the voltage drop across the internal resistance (current of the external circuit times the internal resistance.)

Conservation of Energy & Electrical Efficiency:

Electrical energy is useful to us mostly because it is easy to transport and can be easily converted to or from other forms of energy. Of course, conversion involves wast, typically as heat.

Electrical energy consumed can be determined by multiplying power by time \begin{align*}(E = Pt)\end{align*}. Recall the equations for mechanical and thermal energy/work \begin{align*}(PE=mgh, KE=1/2mv^2, Q=mc \Delta T)\end{align*}. An important idea is the efficiency of an electrical device: the fraction of electrical energy consumed that goes into doing useful work \begin{align*}(W_{out}/E_{in})\end{align*}, expressed as a percentage.

Example:

You use a 100 W electric motor to lift a 10 kg mass 5 m, and it takes 20 s.

The electrical energy consumed is \begin{align*}E_{elec} = P \cdot t = (100 \ W) (20 \ s) = 2000 \ J\end{align*}

The work done is against gravity, so we use \begin{align*}PE = mgh = (10 \ kg) (10 \ m/s^2) (5 \ m) = 500 \ J\end{align*}

The efficiency is \begin{align*}W_{out}/E_{in} = (500 \ J) / (2000 \ J) = 0.25 = 25\% \ \text{efficient}\end{align*}

3 Example Problems for Circuits

1) A circuit is wired up with two resistors in series.

Both resistors are in the same ‘river’, so both have the same current flowing through them. Neither resistor has a direct connection to the power supply so neither has 20V across it. But the combined voltages across the individual resistors add up to 20V.

Question: What is the total resistance of the circuit?

Answer: The total resistance is \begin{align*}R_{total}=R_1+R_2=90 \Omega+10 \Omega=100 \Omega\end{align*}

Question: What is the total current coming out of the power supply?

Answer: Use Ohm’s Law \begin{align*}(V=IR)\end{align*} but solve for current \begin{align*}(I=V/R)\end{align*}.

\begin{align*}I_{total}=\frac{V_{total}}{R_{total}}=\frac{20V}{100\Omega}=0.20A\end{align*}

Question: How much power does the power supply dissipate?

Answer: \begin{align*}P=IV\end{align*}, so the total power equals the total voltage multiplied by the total current. Thus, \begin{align*}P_{total}=I_{total}V_{total}=(0.20A)(20V)=4.0W\end{align*}. So the Power Supply is outputting 4W (i.e. 4 Joules of energy per second).

Question: How much power does each resistor dissipate?

Answer: Each resistor has different voltage across it, but the same current. So, using Ohm’s law, convert the power formula into a form that does not depend on voltage.

\begin{align*}P&=IV=I(IR)=I^2R.\\ P_{90 \Omega} &= I^2_{90\Omega}R_{90\Omega}=(0.2A)^2(90\Omega)=3.6W\\ P_{10 \Omega} &= I^2_{10\Omega}R_{10\Omega}=(0.2A)^2(10\Omega)=0.4W\end{align*}

\begin{align*}^*\end{align*}Note: If you add up the power dissipated by each resistor, it equals the total power outputted, as it should–Energy is always conserved.

Question: How much voltage is there across each resistor?

Answer: In order to calculate voltage across a resistor, use Ohm’s law.

\begin{align*}V_{90\Omega} &= I_{90\Omega}R_{90\Omega}=(0.2A)(90\Omega)=18V\\ V_{10\Omega} &= I_{10\Omega}R_{10\Omega}=(0.2A)(10\Omega)=2V\end{align*}

\begin{align*}^*\end{align*}Note: If you add up the voltages across the individual resistors you will obtain the total voltage of the circuit, as you should. Further note that with the voltages we can use the original form of the Power equation \begin{align*}(P=IV)\end{align*}, and we should get the same results as above.

\begin{align*}P_{90\Omega} &= I_{90\Omega}V_{90\Omega}=(18V)(0.2A)=3.6W\\ P_{10\Omega} &= I_{10\Omega}V_{10\Omega}=(2.0V)(0.2A)=0.4W\end{align*}

2) A circuit is wired up with 2 resistors in parallel.

Both resistors are directly connected to the power supply, so both have the same 20V across them. But they are on different ‘rivers’ so they have different current flowing through them. Lets go through the same questions and answers as with the circuit in series.

Question: What is the total resistance of the circuit?

Answer: The total resistance is \begin{align*}\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{90\Omega}+\frac{1}{10\Omega}=\frac{1}{90\Omega}+\frac{9}{90\Omega}=\frac{10}{90\Omega}\end{align*} thus, \begin{align*}R_{total}=\frac{90\Omega}{10}=9\Omega\end{align*}

\begin{align*}^*\end{align*}Note: Total resistance for a circuit in parallel will always be smaller than smallest resistor in the circuit.

Question: What is the total current coming out of the power supply?

Answer: Use Ohm’s Law \begin{align*}(V=IR)\end{align*} but solve for current \begin{align*}(I=V/R)\end{align*}.

\begin{align*}I_{total}=\frac{V_{total}}{R_{total}}=\frac{20V}{9\Omega}=2.2A\end{align*}

Question: How much power does the power supply dissipate?

Answer: \begin{align*}P=IV\end{align*}, so the total power equals the total voltage multiplied by the total current. Thus, \begin{align*}P_{total}=I_{total}V_{total}=(2.2A)(20V)=44.4W\end{align*}. So the Power Supply outputs 44W (i.e. 44 Joules of energy per second).

Question: How much power is each resistor dissipating?

Answer: Each resistor has different current across it, but the same voltage. So, using Ohm’s law, convert the power formula into a form that does not depend on current. \begin{align*}P=IV=\left(\frac{V}{R}\right) V=\frac{V^2}{R}\end{align*} Substituted \begin{align*}I=V/R\end{align*} into the power formula. \begin{align*}P_{90\Omega}=\frac{V^2_{90\Omega}}{R_{90\Omega}}=\frac{(20V)^2}{90\Omega}=4.4W; P_{10\Omega}=\frac{V^2_{10\Omega}}{R_{10}\Omega}=\frac{(20V)^2}{10\Omega}=40W\end{align*}

\begin{align*}^*\end{align*}Note: If you add up the power dissipated by each resistor, it equals the total power outputted, as it should–Energy is always conserved.

Question: How much current is flowing through each resistor?

Answer: Use Ohm’s law to calculate the current for each resistor.

\begin{align*}I_{90\Omega}=\frac{V_{90\Omega}}{R_{90\Omega}}=\frac{20V}{90\Omega}=0.22A \qquad I_{10\Omega}=\frac{V_{10\Omega}}{R_{10\Omega}}=\frac{20V}{10\Omega}=2.0A\end{align*}

Notice that the \begin{align*}10\Omega\end{align*} resistor has the most current going through it. It has the least resistance to electricity so this makes sense.

\begin{align*}^*\end{align*}Note: If you add up the currents of the individual ‘rivers’ you get the total current of the of the circuit, as you should.

3) A more complicated circuit is analyzed.

Question: What is the total resistance of the circuit?

Answer: In order to find the total resistance we do it in steps (see pictures. First add the \begin{align*}90 \Omega\end{align*} and \begin{align*}10\Omega\end{align*} in series to make one equivalent resistance of \begin{align*}100\Omega\end{align*} (see diagram at right). Then add the \begin{align*}100\Omega\end{align*} to the \begin{align*}10\Omega\end{align*} in parallel to get one resistor of \begin{align*}9.1\Omega\end{align*}. Now we have two resistors in series, simply add them to get the total resistance of \begin{align*}29.1\Omega\end{align*}.

Question: What is the total current coming out of the power supply?

Answer: Use Ohm’s Law \begin{align*}(V=IR)\end{align*} but solve for current \begin{align*}(I=V/R)\end{align*}. \begin{align*}I_{total}=\frac{V_{total}}{R_{total}}=20V/2.91\Omega=0.69\ Amps\end{align*}

Question: What is the power dissipated by the power supply?

Answer: \begin{align*}P=IV\end{align*}, so the total power equals the total voltage multiplied by the total current. Thus, \begin{align*}P_{total}=I_{total}V_{total}=(0.69A)(20V)=13.8W\end{align*}.

Question: How much power is the \begin{align*}20\Omega\end{align*} resistor dissipating?

Answer: The \begin{align*}20\Omega\end{align*} has the full 0.69Amps running through it because it is part of the ‘main river’ (this is not the case for the other resistors because the current splits). \begin{align*}P_{20 \Omega} = I^2_{20 \Omega} R_{20 \Omega} = (0.69A)^2 (20 \Omega) = 9.5W\end{align*}

Question: If these resistors are light bulbs, order them from brightest to least bright.

Answer: The brightness of a light bulb is directly given by the power dissipated. So we could go through each resistor as we did the \begin{align*}20\Omega\end{align*} guy and calculate the power then simply order them. But, we can also think it out. For the guys in parallel the current splits with most of the current going through the \begin{align*}10\Omega\end{align*} path (less resistance) and less going through the \begin{align*}90\Omega+10\Omega\end{align*} path. Well the second path is ten times the resistance of the first, so it will have one tenth of the total current. Thus, there is approximately and 0.069 Amps going through the \begin{align*}90\Omega\end{align*} and \begin{align*}10\Omega\end{align*} path and 0.621Amps going through the \begin{align*}10\Omega\end{align*} path.

\begin{align*}P_{10 \Omega} &= I^2_{10\Omega}R_{10 \Omega}=(0.621A)^2(10\Omega)=3.8W\\ P_{90+10\Omega} &= I^2_{90+10\Omega}R_{90 + 10 \Omega}=(0.069A)^2(100\Omega)=0.5W\end{align*}

We now know that the \begin{align*}20\Omega\end{align*} is the brightest, \begin{align*}10\Omega\end{align*} is second and then the \begin{align*}90\Omega\end{align*} and last the \begin{align*}10\Omega\end{align*} (-these last two have same current flowing through them, so \begin{align*}90\Omega\end{align*} is brighter due to its higher resistance).

\begin{align*}^*\end{align*}Note: Adding up these two plus the 9.5W from the \begin{align*}20\Omega\end{align*} resistor gives us 13.8W, which is the total power previously calculated, so we have confidence everything is good.

Electric Circuits Problem Set

1. The current in a wire is 4.5 A.
1. How many coulombs per second are going through the wire?
2. How many electrons per second are going through the wire?
2. A light bulb with resistance of \begin{align*}80 \ \Omega\end{align*} is connected to a 9 V battery.
1. What is the electric current going through the bulb?
2. What is the power (i.e. wattage) dissipated in this light bulb with the 9 V battery?
3. How many electrons leave the battery every hour?
4. How many joules of energy does the battery provide every hour?
3. A 120 V, 75 W light bulb is shining in your room and you ask yourself...
1. What is the resistance of the light bulb?
2. If you use a 9 V battery, how bright would it shine (i.e. what is its power output)?
4. A bird is standing on an electric transmission line carrying 3000 A of current. A wire like this has about \begin{align*}3.0 \times 10^{-5} \ \Omega\end{align*} of resistance per meter. The bird’s feet are 6 cm apart. The bird, itself, has a resistance of about \begin{align*}4 \times 10^5 \ \Omega\end{align*}.
1. What is the potential difference (i.e. voltage) across the birds two feet?
2. What current goes through the bird?
3. What is the power dissipated by the bird?
4. By how many joules of energy does the bird heat up every hour?
5. Which light bulb will shine brighter? Which light bulb will shine for a longer amount of time? Draw the schematic diagram for both situations. Note that the objects on the right are batteries, not resistors.
6. Regarding the circuit to the right,
1. If the ammeter reads 2 A, what is the voltage?
2. How many watts is the power supply supplying?
3. How many watts are dissipated in each resistor?
7. Three \begin{align*}82 \ \Omega\end{align*} resistors and one \begin{align*}12 \ \Omega\end{align*} resistor are wired in parallel with a 9 V battery.
1. Draw the schematic diagram.
2. What is the total resistance of the circuit?
8. What will the ammeter read for the circuit shown to the right?
9. Draw the schematic of the circuit below.
10. What does the ammeter read and which resistor is dissipating the most power?
11. Analyze the circuit to the right.
1. Find the current going out of the power supply.
2. How many joules per second of energy is the power supply giving out?
3. Find the current going through the \begin{align*}75 \ \Omega\end{align*} light bulb.
4. Find the current going through the \begin{align*}50 \ \Omega\end{align*} light bulbs (hint: it’s the same, why?).
5. Order the light bulbs in terms of brightness.
6. If they were all wired in parallel, order them in terms of brightness.
12. Find the total current output by the power supply and the power dissipated by the \begin{align*}20 \ \Omega\end{align*} resistor.
13. You have a 600 V power source, two \begin{align*}10 \ \Omega\end{align*} toasters that both run on 100 V, and a single \begin{align*}25 \ \Omega\end{align*} resistor.
1. Show with a schematic diagram how you would wire them up so the toasters run properly.
2. What is the power dissipated by the toasters?
3. Show where you could put a fuse to make sure the toasters don’t draw more than 15 amps.
14. Given three resistors, \begin{align*}200 \ \Omega, 300 \ \Omega\end{align*} and \begin{align*}600 \ \Omega\end{align*} and a 120 V power source connect them in a way to heat a container of water as rapidly as possible.
1. Draw the circuit diagram.
2. How many joules of heat are generated after 5 minutes?
15. The useful work a light bulb does is emitting light (duh). The rest is “wasted” as heat.
1. If a standard incandescent 60 W bulb is on for 1 minute and generates 76 J of light energy, what is its efficiency? How much heat energy is produced?
2. If the efficiency of a CFL (compact fluorescent) bulb is 20%, how many joules of light energy will a 60 W bulb produce?
16. Most microwave ovens are 1000 W devices. You heat 1 cup (8 fl. oz., or 30 ml) of room temperature water \begin{align*}(20^\circ C)\end{align*} for 30 seconds in a microwave.
1. What current does the oven draw?
2. If the water heats up to \begin{align*}90^\circ C\end{align*}, what is the heating efficiency of the oven?
17. Look at the following scheme of four identical light bulbs connected as shown. Answer the questions below giving a justification for your answer:
1. Which of the four light bulbs is/are the brightest?
2. Which light bulbs is/are the dimmest?
3. Tell in the following cases which other light bulbs go out if:
1. bulb \begin{align*}A\end{align*} goes out
2. bulb \begin{align*}B\end{align*} goes out
3. bulb \begin{align*}D\end{align*} goes out
4. Tell in the following cases which other light bulbs get dimmer, and which get brighter if:
1. bulb \begin{align*}B\end{align*} goes out
2. bulb \begin{align*}D\end{align*} goes out
18. In the circuit shown here, the battery produces an emf of 1.5 V and has an internal resistance of \begin{align*}0.5 \ \Omega\end{align*}.
1. Find the total resistance of the external circuit.
2. Find the current drawn from the battery.
3. Determine the terminal voltage of the battery.
4. Show the proper connection of an ammeter and a voltmeter that could measure voltage across and current through the \begin{align*}2 \ \Omega\end{align*} resistor. What would these instruments read?
19. Students design a circuit with a single unknown resistor, making the following measurements:
Voltage (V) Current (A)
15 0.11
12 0.080
10 0.068
8.0 0.052
6.0 0.040
4.0 0.025
2.0 0.010

(a) Show a circuit diagram with the connections to the power supply, ammeter and voltmeter.

(b) Graph voltage vs. current either here by hand or on the computer; find the linear best-fit.

(c) Use this line to determine the resistance.

(d) How confident can you be of the results?

(e) Use the graph to determine the current if the voltage were 13 V.

1. The 2010 Toyota Prius \begin{align*}(m=1250 \ kg)\end{align*} battery is rated at 201.6 V with a capacity of 6.5 Ah.
1. What is the total energy stored in this battery on a single charge?
2. If you used the battery alone to accelerate to 65 mph one time (assuming no friction), what percentage of the battery capacity would you use?
2. A certain 48-V electric forklift can lift up to 7000 lb at a maximum rate of 76 ft/min.
1. What is its power?
2. What current must the battery produce to achieve this power?

1. 4.5C
2. \begin{align*}2.8 \times 10^{19}\end{align*} electrons
1. 0.11 A
2. 1.0 W
3. \begin{align*}2.5 \times 10^{21}\end{align*} electrons
4. 3636 W
1. \begin{align*}192 \ \Omega\end{align*}
2. 0.42 W
1. 5.4 mV
2. \begin{align*}1.4 \times 10^{-8} \ A\end{align*}
3. \begin{align*}7.3 \times 10^{-11} \ W\end{align*}, not a lot
4. \begin{align*}2.6 \times 10^{-7} \ J\end{align*}
1. left = brighter, right = longer
1. 224 V
2. 448 W
3. 400 W by \begin{align*}100 \ \Omega\end{align*} and 48 W by \begin{align*}12 \ \Omega\end{align*}
2. (b) \begin{align*}8.3 \ \Omega\end{align*}
3. 0.5A
1. 0.8A and the \begin{align*}50 \ \Omega\end{align*} on the left
1. 0.94 A
2. 112 W
3. 0.35 A
4. 0.94 A
5. 50, 45, 75 \begin{align*}\Omega\end{align*}
6. dimmer; total resistance increases
7. \begin{align*}45 \ \Omega\end{align*}
1. 0.76 A
2. 7.0 W
2. (b) 1000 W
3. discuss in class
1. 2.1%, 3524 J
2. 720 J
1. 8.3A
2. 29%
1. lightbulb \begin{align*}A\end{align*}
2. lightbulbs \begin{align*}B,C\end{align*}
1. All
2. C
3. None
1. \begin{align*}A\end{align*} dimmer, \begin{align*}D\end{align*} brighter, \begin{align*}C\end{align*} out
2. \begin{align*}A\end{align*} dimmer, \begin{align*}B,C\end{align*} brighter
1. \begin{align*}3.66 \ \Omega\end{align*}
2. 0.36 A
3. 1.32 V
1. \begin{align*}4.7 \times 10^6 \ J\end{align*}
2. 11.2%
1. 12300 W
2. 256 A

OPTIONAL PROBLEM:

1. Refer to the circuit diagram below and answer the following questions.
1. What is the resistance between \begin{align*}A\end{align*} and \begin{align*}B\end{align*}?
2. What is the resistance between \begin{align*}C\end{align*} and \begin{align*}B\end{align*}?
3. What is the resistance between \begin{align*}D\end{align*} and \begin{align*}E\end{align*}?
4. What is the the total equivalent resistance of the circuit?
5. What is the current leaving the battery?
6. What is the voltage drop across the \begin{align*}12 \ \Omega\end{align*} resistor?
7. What is the voltage drop between \begin{align*}D\end{align*} and \begin{align*}E\end{align*}?
8. What is the voltage drop between \begin{align*}A\end{align*} and \begin{align*}B\end{align*}?
9. What is the current through the \begin{align*}25 \ \Omega\end{align*} resistor?
10. What is the total energy dissipated in the \begin{align*}25 \ \Omega\end{align*} if it is in use for 11 hours?

1. (a) \begin{align*}9.1 \ \Omega\end{align*} (b) \begin{align*}29.1 \ \Omega\end{align*} (c) \begin{align*}10.8 \ \Omega\end{align*} (d) \begin{align*}26.8 \ \Omega\end{align*} (e) 1.8 A (f) 21.5 V (g) 19.4 V (h) 6.1 V (i) 0.24 A (j) 16 kW

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