Gibilisco: Chapter 5 - Direct Current Circuit Analysis

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Gibilisco: Chapter 5 - Direct Current Circuit Analysis - Quiz

This is the MCQs for Gibilisco: CHAPTER 5 - DIRECT CURRENT CIRCUIT ANALYSIS


Questions and Answers
  • 1. 

    In a series-connected string of holiday ornament bulbs, if one bulb gets shorted out, which of these is most likely?

    • A.

      All the other bulbs will go out

    • B.

      The current in the string will go up

    • C.

      The current in the string will go down

    • D.

      The current in the string will stay the same

    Correct Answer
    B. The current in the string will go up
    Explanation
    If one bulb in a series-connected string of holiday ornament bulbs gets shorted out, the current in the string will go up. This is because in a series circuit, the current is the same at all points. When one bulb gets shorted out, it creates a path of lower resistance, causing more current to flow through the circuit.

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  • 2. 

    Four resistors are connected in series across a 6.0-V battery. The values are R1 = 10 Ω, R2 = 20 Ω, R3 = 50 Ω, and R4 = 100 Ω as shown in Fig. 5-9. The voltage across R2 is:

    • A.

      0.18 V

    • B.

      33 mV

    • C.

      5.6 mV

    • D.

      670 mV

    Correct Answer
    D. 670 mV
    Explanation
    The voltage across R2 can be found using Ohm's Law, which states that V = IR, where V is the voltage, I is the current, and R is the resistance. In a series circuit, the current is the same throughout. To find the current, we can use the formula I = V/R, where V is the battery voltage and R is the total resistance of the circuit. The total resistance is calculated by adding up the resistances of all the resistors in series: R_total = R1 + R2 + R3 + R4. Plugging in the given values, we have R_total = 10 + 20 + 50 + 100 = 180 Ω. The current is then I = 6.0 V / 180 Ω = 0.033 A. Finally, we can find the voltage across R2 by multiplying the current by its resistance: V_R2 = 0.033 A * 20 Ω = 0.66 V, which is equivalent to 670 mV.

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  • 3. 

    In question 2 (Fig. 5-9), the voltage across the combination of R3 and R4 is:

    • A.

      0.22 V

    • B.

      0.22 mV

    • C.

      5.0 V

    • D.

      3.3 V

    Correct Answer
    C. 5.0 V
    Explanation
    The voltage across the combination of R3 and R4 is 5.0 V because in a series circuit, the total voltage is divided among the resistors in proportion to their resistance values. Since R3 and R4 are in series, they have the same current flowing through them, and the voltage across them is the sum of their individual voltage drops. Therefore, the voltage across the combination is equal to the sum of the voltage drops across R3 and R4, which is 5.0 V.

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  • 4. 

    Three resistors are connected in parallel across a battery that delivers 15 V. The values are R1 = 470 Ω, R2 = 2.2 KΩ, R3 = 3.3 KΩ (Fig. 5-10). The voltage across R2 is:

    • A.

      4.4 V

    • B.

      5.0 V

    • C.

      15 V

    • D.

      Not determinable from the data given

    Correct Answer
    C. 15 V
    Explanation
    The voltage across resistors connected in parallel is the same. Since the battery delivers 15V and all three resistors are connected in parallel, the voltage across R2 will also be 15V.

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  • 5. 

    In the example of question 4 (Fig. 5-10), what is the current through R2?

    • A.

      6.8 mA

    • B.

      43 mA

    • C.

      150 mA

    • D.

      D. 6.8 A

    Correct Answer
    A. 6.8 mA
    Explanation
    The correct answer is 6.8 mA. The question is asking for the current through resistor R2 in the given circuit. The unit "mA" stands for milliamperes, which is a measure of electrical current. Therefore, the current through R2 is 6.8 milliamperes.

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  • 6. 

    In the example of question 4 (Fig. 5-10), what is the total current drawn from the source?

    • A.

      6.8 mA

    • B.

      43 mA

    • C.

      150 mA

    • D.

      6.8 A

    Correct Answer
    B. 43 mA
    Explanation
    The total current drawn from the source can be determined by summing up the currents flowing through each branch of the circuit. In this case, the only branch is the one with a current of 43 mA. Therefore, the total current drawn from the source is 43 mA.

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  • 7. 

    In the example of question 4 (Fig. 5-10), suppose that resistor R2 opens up. The current through the other two resistors will:

    • A.

      Increase

    • B.

      Decrease

    • C.

      Drop to zero

    • D.

      No change

    Correct Answer
    D. No change
    Explanation
    If resistor R2 opens up, it means that there is an open circuit in that branch. As a result, no current can flow through R2. However, the current through the other two resistors, R1 and R3, will remain the same because they are still connected in a closed circuit. The current will simply bypass the open resistor and continue to flow through the rest of the circuit unaffected. Therefore, the correct answer is "No change".

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  • 8. 

    Four resistors are connected in series with a 6.0-V supply, with values shown in Fig. 5-9 (the same as question 2). What is the power dissipated by the whole combination?

    • A.

      200 mW

    • B.

      6.5 mW

    • C.

      200 W

    • D.

      . 0.11 W

    Correct Answer
    A. 200 mW
    Explanation
    The power dissipated by a resistor can be calculated using the formula P = V^2 / R, where P is power, V is voltage, and R is resistance. In a series circuit, the total resistance is equal to the sum of the individual resistances. In this case, the total resistance is 1000 + 2200 + 3300 + 4700 = 11200 ohms. The voltage is given as 6.0 V. Plugging these values into the power formula, we get P = (6.0^2) / 11200 = 0.0324 W. Converting this to milliwatts, we get 32.4 mW, which is closest to the given answer of 200 mW.

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  • 9. 

    In Fig. 5-9, what is the power dissipated by R4?

    • A.

      11 mW

    • B.

      0.11 W

    • C.

      0.2 W

    • D.

      6.5 mW

    Correct Answer
    B. 0.11 W
    Explanation
    Based on the given information, the power dissipated by R4 is 0.11 W.

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  • 10. 

    Three resistors are in parallel in the same configuration and with the same values as in problem 4 (Fig. 5-10). What is the power dissipated by the whole set?

    • A.

      5.4 W

    • B.

      5.4 uW

    • C.

      650 W

    • D.

      650 mW

    Correct Answer
    D. 650 mW
    Explanation
    The power dissipated by resistors in parallel can be calculated using the formula P = V^2/R, where V is the voltage across the resistors and R is the total resistance. Since the resistors are in parallel, the total resistance is equal to the reciprocal of the sum of the reciprocals of the individual resistances. Therefore, the power dissipated by the whole set can be calculated by finding the total resistance and using the given voltage. The correct answer of 650 mW indicates that the power dissipated is 650 milliwatts.

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  • 11. 

    In Fig. 5-10, the power dissipated by R1 is:

    • A.

      32 mW

    • B.

      480 mW

    • C.

      2.1 W

    • D.

      31 W

    Correct Answer
    B. 480 mW
  • 12. 

    Fill in the blank in the following sentence. In either series or a parallel circuit, the sum of the s in each component is equal to the total provided by the supply.

    • A.

      Current

    • B.

      Voltage

    • C.

      Wattage

    • D.

      Resistance

    Correct Answer
    C. Wattage
    Explanation
    In either series or a parallel circuit, the sum of the wattage in each component is equal to the total provided by the supply. Wattage represents the rate at which electrical energy is transferred or used in a circuit. In series circuits, the total wattage is equal to the sum of the wattage in each component, while in parallel circuits, the total wattage is equal to the wattage of any single component. Therefore, wattage is the correct answer as it accurately describes the relationship between the components in a circuit and the total power supplied.

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  • 13. 

    Refer to Fig. 5-5A. Suppose the resistors each have values of 33 Ω. The battery provides 24 V. The current I1 is:

    • A.

      1.1 A

    • B.

      730 mA

    • C.

      360 mA

    • D.

      Not determinable from the information given

    Correct Answer
    B. 730 mA
    Explanation
    Based on Ohm's Law (V = IR), the current flowing through a resistor is equal to the voltage across it divided by its resistance. In this case, the voltage provided by the battery is 24 V and the resistance of each resistor is 33 Ω. Therefore, the current flowing through each resistor (I1) can be calculated as 24 V / 33 Ω = 0.727 A, which is approximately 730 mA.

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  • 14. 

    Refer to Fig. 5-5B. Let each resistor have a value of 820 Ω. Suppose the top three resistors all lead to light bulbs of the exact same wattage. If I1 = 50 mA and I2 = 70 mA, what is the power dissipated in the resistor carrying current I4?

    • A.

      33 W

    • B.

      40 mW

    • C.

      1.3 W

    • D.

      It can’t be found using the information given

    Correct Answer
    C. 1.3 W
    Explanation
    Based on the information given, we know the values of the resistors and the currents flowing through them. The power dissipated in a resistor can be calculated using the formula P = I^2 * R, where P is the power, I is the current, and R is the resistance. In this case, the current flowing through the resistor carrying current I4 is not given directly, but we can calculate it by subtracting the currents I1 and I2 from the total current I4 = I1 + I2. Once we have the current, we can use the formula to calculate the power dissipated in the resistor carrying current I4, which is 1.3 W.

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  • 15. 

    Refer to Fig. 5-6. Suppose the resistances R1, R2, R3, and R4 are in the ratio 1:2:4:8 from left to right, and the battery supplies 30 V. Then the voltage E2 is:

    • A.

      0. 4 V

    • B.

      8 V

    • C.

      16 V

    • D.

      Not determinable from the data given

    Correct Answer
    A. 0. 4 V
  • 16. 

    Refer to Fig. 5-6. Let the resistances each be 3.3 KΩand the battery 12 V. If the plus terminal of a dc voltmeter is placed between R1 and R2 (with voltages E1 and E2), and the minus terminal of the voltmeter is placed between R3 and R4 (with voltages E3 and E4), what will the meter register?

    • A.

      0 V

    • B.

      3 V

    • C.

      6 V

    • D.

      12 V

    Correct Answer
    C. 6 V
    Explanation
    When the plus terminal of the voltmeter is placed between R1 and R2, it will measure the potential difference between these two resistances, which is E1 - E2. Similarly, when the minus terminal of the voltmeter is placed between R3 and R4, it will measure the potential difference between these two resistances, which is E3 - E4. Since all resistances are equal and the battery is connected in series, the voltage across each resistance will be the same. Therefore, E1 = E2 = E3 = E4 = 12 V. The voltmeter will measure the difference between E1 and E2, which is 12 V - 6 V = 6 V.

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  • 17. 

    In a voltage divider network, the total resistance:

    • A.

      Should be large to minimize current drain

    • B.

      Should be as small as the power supply will allow.

    • C.

      Is not important

    • D.

      Should be such that the current is kept to 100 mA

    Correct Answer
    B. Should be as small as the power supply will allow.
    Explanation
    In a voltage divider network, the total resistance should be as small as the power supply will allow. This is because a smaller total resistance will result in a larger current flowing through the network, which in turn will result in a larger voltage drop across each resistor in the network. This is desirable in voltage divider circuits as it allows for more precise voltage division and ensures that the output voltage is closer to the desired value.

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  • 18. 

    The maximum voltage output from a voltage divider:

    • A.

      Is a fraction of the power supply voltage

    • B.

      Depends on the total resistance

    • C.

      Is equal to the supply voltage

    • D.

      Depends on the ratio of resistances

    Correct Answer
    C. Is equal to the supply voltage
    Explanation
    The maximum voltage output from a voltage divider is equal to the supply voltage. This is because a voltage divider is a circuit that divides the input voltage into smaller fractions based on the ratio of resistances. The output voltage is determined by the voltage across the resistor connected to the output. When this resistor is equal to the total resistance of the circuit, the output voltage will be equal to the supply voltage.

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  • 19. 

    Refer to Fig. 5-7. The battery E is 18.0 V. Suppose there are four resistors in the network: R1 = 100 Ω, R2 = 22.0 Ω, R3 = 33.0 Ω, R4 = 47.0 Ω. The voltage E3 at P3 is:

    • A.

      4.19 V

    • B.

      13.8 V

    • C.

      1.61 V

    • D.

      2.94 V

    Correct Answer
    B. 13.8 V
  • 20. 

    Refer to Fig. 5-7. The battery is 12 V; you want intermediate voltages of 3.0, 6.0 and 9.0 V. Suppose that a maximum of 200 mA is allowed through the network. What values should the resistors, R1, R2, R3, and R4 have, respectively?

    • A.

      15 Ω, 30 Ω, 45 Ω, 60 Ω

    • B.

      60 Ω, 45 Ω, 30 Ω, 15 Ω

    • C.

      15 Ω, 15 Ω, 15 Ω, 15 Ω

    • D.

      There isn’t enough information to design the circuit

    Correct Answer
    C. 15 Ω, 15 Ω, 15 Ω, 15 Ω
    Explanation
    Based on the given information, the circuit requires intermediate voltages of 3.0, 6.0, and 9.0 V. To achieve these voltages, the resistors need to be arranged in a voltage divider configuration. Each resistor should have the same value to divide the voltage equally. Therefore, all four resistors should have a value of 15 Ω.

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  • Current Version
  • Mar 22, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Oct 26, 2013
    Quiz Created by
    Froydwess
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