Ahot water heating system has developed length of 200 feet of 1-inch type m tubing. in this circuit there are soldered fittings, also 1 inch in size, as follows: 60 standard 90° ells, 10 standard 45° ells and 4 gate valves. calculate the total equivalent length of the system.

1. Resistance

2. Store chmical energy and transfer it to electrical energy when a circuit is connected

3.Electric current

4. Series and parallel

5. can't see diagram

6. series

7. Parallel

8. can't not see diagram

Explanation:

1. Resistance

2. Store chemical energy and transfer it to electrical energy when a circuit is connected

3. Electric current

4. Series and parallel

5. cell (battery)

6. series

7. Parallel

8. Diagram B

9. 12Ω resistor

10. protect equipment by stopping the flow of electric current; protect your house from fire.

11. GFCI outlets prevent electrocution if you are touching a wet appliance. GFCI outlets are found in wet areas.

The person that answered before me is right except for question 1

1. 5.9 Ω

2. 0.72 A

3. 2.4 V

4. R1

5. RC circuit

6. 2.0 W

7. R4

Explanation:

I took the test and I got question 1 wrong. (K12 Physics sem 2 8.05)

7.33 Ω

Explanation:

10Ω resistor and a 5 Ω are connected in parallel

so,

R₁=(10X5)/(10+5)

=(50/15) Ω

=3.33 Ω

This pair is then connected in series in with another 4 Ω resistor

R₂=R₁+4

=(3.33+4) Ω

=7.33 Ω

resistance of 4 ohm, 10 ohm and 7 ohm is connected in parallel.

so net resistance in parallel is given by

[tex]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}[/tex]

[tex]\frac{1}{R} = \frac{1}{4} + \frac{1}{10} + \frac{1}{7}[/tex]

[tex]R = \frac{140}{69} ohm[/tex]

now it is connected to 6 ohm resistance in series

so net resistance is given by

[tex]R_{net} = R_1 + R_2[/tex]

[tex]R_{net} = 6 ohm + \frac{140}{69} ohm[/tex]

[tex]R_{net} = 8.03 ohm[/tex]

1) Let's start by calculating the equivalent resistance of the three resistors in parallel, [tex]R_2, R_3, R_4[/tex]:

[tex]\frac{1}{R_{234}}= \frac{1}{R_2}+ \frac{1}{R_3}+ \frac{1}{R_4}= \frac{1}{4.5 \Omega}+ \frac{1}{1.3 \Omega}+ \frac{1}{6.3 \Omega}=1.15 \Omega^{-1}[/tex]

From which we find

[tex]R_{234}= \frac{1}{1.15 \Omega^{-1}}=0.9 \Omega[/tex]

Now all the resistors are in series, so the equivalent resistance of the circuit is the sum of all the resistances:

[tex]R_{eq}=R_1 + R_{234} = 5 \Omega + 0.9 \Omega = 5.9 \Omega[/tex]

So, the correct answer is D)

2) Let's start by calculating the equivalent resistance of the two resistors in parallel:

[tex]\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}[/tex]

From which we find

[tex]R_{23} = 2.5 \Omega[/tex]

And these are connected in series with a resistor of [tex]10 \Omega[/tex], so the equivalent resistance of the circuit is

[tex]R_{eq}=10 \Omega + 2.5 \Omega = 12.5 \Omega[/tex]

And by using Ohm's law we find the current in the circuit:

[tex]I= \frac{V}{R_{eq}}= \frac{9 V}{12.5 \Omega}=0.72 A[/tex]

So, the correct answer is C).

3) Let' start by calculating the equivalent resistance of the two resistors in parallel:

[tex]\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{5 \Omega}+ \frac{1}{5 \Omega}= \frac{2}{5 \Omega}[/tex]

From which we find

[tex]R_{23} = 2.5 \Omega[/tex]

Then these are in series with all the other resistors, so the equivalent resistance of the circuit is

[tex]R_{eq}=R_1 + R_{23}+R_4 = 5 \Omega + 2.5 \Omega + 5 \Omega =12.5 \Omega[/tex]

And by using Ohm's law we find the current flowing in the circuit:

[tex]I= \frac{V}{R_{eq}}= \frac{12 V}{12.5 \Omega}=0.96 A[/tex]

And so the voltage read by the voltmeter V1 is the voltage drop across the resistor 2-3:

[tex]V= I R_{23} = (0.96 A)(2.5 \Omega)=2.4 V[/tex]

So, the correct answer is D).

4) Again, let's start by calculating the equivalent resistance of the two resistors in parallel:

[tex]\frac{1}{R_{23}} = \frac{1}{R_2}+ \frac{1}{R_3}= \frac{1}{13 \Omega}+ \frac{1}{18 \Omega}=0.13 \Omega^{-1}[/tex]

From which we find

[tex]R_{23} = 7.55 \Omega[/tex]

Now all the resistors are in series, so the equivalent resistance of the circuit is:

[tex]R_{eq}= R_1 + R_{23}+R_4=8.5 \Omega+7.55 \Omega + 3.2 \Omega = 19.25 \Omega[/tex]

The current in the circuit is given by Ohm's law

[tex]I= \frac{V}{R_{Eq}}= \frac{15 V}{19.25 \Omega}=0.78 A[/tex]

Now we can compare the voltage drops across the resistors. Resistor 1:

[tex]V_1 = I R_1 = (0.78 A)(8.5 \Omega)=6.63 V[/tex]

Resistor 2 and resistor 3 are in parallel, so they have the same voltage drop:

[tex]V_2 = V_3 = V_{23} = I R_{23} = (0.78 A)(7.55 \Omega)=5.89 V[/tex]

Resistor 4:

[tex]V_4 = I R_4 = (0.78 A)(3.2 \Omega)=2.50 V[/tex]

So, the greatest voltage drop is on resistor 1, so the correct answer is D).

5) the figure shows a circuit with a resistor R and a capacitor C, so it is an example of RC circuit. Therefore, the correct answer is D).

6) The circuit is the same as part 4), so the calculations are exactly the same. Therefore, the power dissipated on resistor 3 is

[tex]P_3 = I_3^2 R_3 = \frac{V_3^2}{R_3}= \frac{(5.89 V)^2}{18 \Omega}=2.0 W[/tex]

So, correct answer is B).

7) The circuit is the same as part 4), so we can use exactly the same calculation, and we immediately see that the resistor with lowest voltage drop was R4 (2.50 V), so the correct answer is B) R4.

(3) 4.62 ohms

Explanation:

Te equivalent resistance R is gotten from

1/R = 1/R₁ + 1/R₂ + 1/R₃ where R₁ = 10.0 ohm, R₂ = 15.0 ohm and R₃ = 20.0 ohm

1/R = 1/10 + 1/15 + 1/20

1/R = 0.1 + 0.0667 + 0.05

1/R = 0.2167

R = 1/0.2167

R = 4.62 ohms

1. Electric current

2. In fact it is electrons that flow through a metallic conductor, and they flow from the negative terminal to the positive terminal

3. Series and parallel

4. Series

5. Parallel

6.

7.

8.A fuse is a small, thin conductor designed to melt and separate into two pieces for the purpose of breaking a circuit in the event of excessive current. A circuit breaker is a specially designed switch that automatically opens to interrupt circuit current in the event of an overcurrent condition.

9.

10. The power of batteries in a circuit. The key functions of a battery and bulb in a circuit are explained. A battery is a source of energy which provides a push - a voltage - of energy to get the current flowing in a circuit. A bulb uses the electrical energy provided by the battery, but does not use current.

hope this helps

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