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1) What is the equivalent resistance of the circuit shown?
A) 1.2 Ú
B) 6.2 Ú
C) 0.87 Ú
D) 5.9 Ú

2) One 10.0 Ω resistor is wired in series with two 5.0 Ω resistors in parallel. A 9.0 V battery supplies power to the circuit. What is the total current through the circuit?
A) 0.45 A
B) 1.1 A
C) 0.72 A
D) 0.87 A

3) If each resistor in the circuit shown is 5.0 Ω and the battery is 12.0 V, what does the first voltmeter (V1) read?
A) 0.35 V
B) 0.96 V
C) 0.12 V
D) 2.4 V

4) In the circuit shown, the greatest voltage drop will occur across which of the following resistors?
A) R2
B) R3
C) R4
D) R1

5) The circuit shown is an example of which of the following?
A) power circuit
B) RL circuit
C) voltage divider
D) RC circuit

6) For the circuit shown, what is the power dissipated by R3?
A) 5.2 W
B) 2.0 W
C) 2.5 W
D) 2.8 W

7) In the circuit shown, the lowest voltage drop will occur across which of the following resistors?
A) R1
B) R4
C) R2
D) R3
(Pictures below in order. question 1, 3, 4& 6, 5, and 7)

Please help with Physics Circuits Only answer if you know no spam pleaseಠ益ಠ 1 What is the equivalent resistance of the circuit shown A 12 Ú B 62 Ú C 087 Ú D 59 class=
Please help with Physics Circuits Only answer if you know no spam pleaseಠ益ಠ 1 What is the equivalent resistance of the circuit shown A 12 Ú B 62 Ú C 087 Ú D 59 class=
Please help with Physics Circuits Only answer if you know no spam pleaseಠ益ಠ 1 What is the equivalent resistance of the circuit shown A 12 Ú B 62 Ú C 087 Ú D 59 class=
Please help with Physics Circuits Only answer if you know no spam pleaseಠ益ಠ 1 What is the equivalent resistance of the circuit shown A 12 Ú B 62 Ú C 087 Ú D 59 class=
Please help with Physics Circuits Only answer if you know no spam pleaseಠ益ಠ 1 What is the equivalent resistance of the circuit shown A 12 Ú B 62 Ú C 087 Ú D 59 class=

Respuesta :

skyluke89
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.

The current in the given circuits are obtained from the value of the voltage

source and the equivalent resistance of the circuit.

Responses:

  • 1) 5.9 Ω
  • 2) C) 0.72 A
  • 3) D) 2.4 V
  • 4) A) R₁
  • 5) RC circuit
  • 6) B) 2.0 W
  • 7) B) R₄

Methods by which the above responses are obtained;

1) The equivalent resistance is given by addition of the series and parallel resistances as follows;

[tex]\displaystyle R_{e} = \mathbf{ R_1 + \frac{1}{\frac{1}{R_2} + \frac{1}{R_3} +\frac{1}{R_4} }}[/tex]

Which gives;

[tex]\displaystyle R_{e} = 5 + \frac{1}{\frac{1}{4.5} + \frac{1}{1.3} +\frac{1}{6.3} } =\frac{1843}{314} \approx 5.9[/tex]

The equivalent resistance, [tex]R_{equivalent}[/tex] ≈ 5.9 Ω

  • The equivalent resistance of the circuit is D) 5.9 Ω

2) The equivalent resistor of the circuit, [tex]\mathbf{R_e}[/tex], is given as follows;

[tex]\displaystyle R_e = 10 \, \Omega + \frac{1}{\frac{1}{5 \, \Omega} +\frac{1}{5 \, \Omega} } = 12.5 \, \Omega[/tex]

The current, I, in the circuit is therefore;

[tex]\displaystyle I = \frac{9.0 \, V}{12.5 \, \Omega } = \mathbf{0.72 \, A}[/tex]

  • The current in the circuit is C) 0.72 A

3) R₁ = R₂ = R₃ = R₄ = 5.0 Ω

[tex]\displaystyle R_{e} = 5 + \frac{1}{\frac{1}{5} + \frac{1}{5}} + 5 = 12.5[/tex]

[tex]R_{e}[/tex] = 12.5 Ω

[tex]\displaystyle I_T = \frac{12.0 \, V}{12.5 \, \Omega} = \mathbf{0.96 \, A}[/tex]

The voltage reading of the first voltmeter is therefore;

[tex]\displaystyle v_1 = 0.96 \, A \times \frac{1}{\frac{1}{5 \, \Omega} + \frac{1}{5 \, \Omega} } = 2.4 \, V[/tex]

  • The voltage reading of the first voltmeter, v₁ is D) 2.4 V

4) The current in the circuit, [tex]\mathbf{I_T}[/tex], is found as follows;

[tex]\displaystyle R_e = 8.5 + \frac{1}{\frac{1}{13} + \frac{1}{18} } + 3.2 = \frac{5967}{310}[/tex]

[tex]\displaystyle I_T = \frac{15 \, V}{\frac{5,967}{310} \, \Omega } = \mathbf{\frac{1,550}{1,989} \, A}[/tex]

By current divider rule, we have;

[tex]\displaystyle I_3 = \frac{5,967}{1,989} \times \frac{13}{13+ 18} = \mathbf{\frac{50}{153}}[/tex]

[tex]\displaystyle Voltage} \ drop \ across \ R_1 = \frac{1,550}{1,989} \, A \times 8.5 \, \Omega \approx \mathbf{6.623 \, V}[/tex]

[tex]\displaystyle Voltage \ drop \ across \ \mathbf{R_3}, \, v_3 = \frac{50}{153} \, A \times 18 \, \Omega \approx 5.88 \, V[/tex]

By calculation, we have;

[tex]\displaystyle Voltage \ drop \ across \ \mathbf{R_2}, \, v_2 = \frac{100}{221} \, A \times 13 \, \Omega \approx 5.88 \, V = Voltage \ drop \ across \ R_3[/tex]

[tex]\displaystyle Voltage \ drop \ across \ \mathbf{R_4}, \, v_4 = \frac{5,967}{1,989} \, A \times 3.2 \, \Omega \approx 2.49 \, V[/tex]

Therefore;

  • The greatest voltage drop occurs across the resistor D) R₁

5) The component in the circuit are a voltage source, resistor, and capacitor, therefore, the circuit is a D) RC circuit.

6) The power dissipated by R₃, is given by P₃ = I₃² × R₃

Therefore;

[tex]\displaystyle P_3 = \left(\frac{50}{153} \, A\right)^2 \times 18 \, \Omega \approx 2.0 \, W[/tex]

  • The power dissipated by R₃, P₃ is approximately, B) 2.0 W

7) The lowest voltage drop (2.49 V) occurs across R₄, as calculated in question (4) above;

[tex]\displaystyle Voltage \ drop \ across \ R_4 = \frac{5,967}{1,989} \, A \times 3.2 \, \Omega \approx 2.49 \, V[/tex]

  • The lowest voltage drop occurs across [tex]\underline{B) \ R_4}[/tex]

Learn more about series and parallel circuits here:

https://brainly.com/question/869403

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