Current Electricity! Play This Trivia Quiz

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| By Tanmay Shankar
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Tanmay Shankar
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Current Electricity! Play This Trivia Quiz - Quiz

Electricity is a form of energy that is obtained from mechanical, solar just to mention but a few of the many energy sources. The electricity moves in currents through wires. The quiz below is this subject. All the best.


Questions and Answers
  • 1. 

    Given three resistors, how many different combinations of these three resistances can be made   

    • A.

      Three

    • B.

      Four

    • C.

      Five

    • D.

      Six

    Correct Answer
    B. Four
    Explanation
    There are four different combinations of these three resistances that can be made. This is because each resistor can either be included or excluded from the combination, resulting in 2^3 = 8 possible combinations. However, two of these combinations have all three resistors included (3 resistors together), and one combination has no resistors included (0 resistors together). Therefore, the remaining four combinations have different numbers of resistors included, resulting in four different combinations.

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  • 2. 

    In the Bohr’s model of hydrogen atom, the electron moves around the nucleus in circular orbit of radius 5 ⤬ 10–11 m. Its time period is 1.5 ⤬ 10–16 s. The current associated with electron motion is

    • A.

      Zero

    • B.

      1.6 ⤬ 10–19 A

    • C.

      1.07 ⤬ 10–3 A

    • D.

      0.17 A

    Correct Answer
    C. 1.07 ⤬ 10–3 A
    Explanation
    In the Bohr's model of the hydrogen atom, the electron moves in a circular orbit around the nucleus. The time period of the electron's motion is given as 1.5 ⤬ 10–16 s. The current associated with electron motion is given by the equation I = q/T, where q is the charge and T is the time period. Since the charge of an electron is 1.6 ⤬ 10–19 C, we can substitute the values into the equation to find the current. Therefore, the current associated with electron motion is 1.6 ⤬ 10–19 C / 1.5 ⤬ 10–16 s = 1.07 ⤬ 10–3 A.

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  • 3. 

    The temperature (T) dependence of resistivity (p) of a semiconductor is represented by

    Correct Answer
    B.
  • 4. 

    Resistance n, each of r Ω, when connected in parallel give an equivalent resistance of R Ω. If these resistances were connected in series, the combination would have a resistance in Ω, equal to

    • A.

      N2R

    • B.

      R / n2

    • C.

      R / n

    • D.

      NR

    Correct Answer
    A. N2R
    Explanation
    When resistances are connected in parallel, the equivalent resistance is given by the formula 1/R = 1/r1 + 1/r2 + ... + 1/rn. In this case, each resistance is r, so the equivalent resistance is 1/R = 1/r + 1/r + ... + 1/r (n times) = n/r. Solving for R, we get R = r/n.

    Now, if the resistances were connected in series, the equivalent resistance would simply be the sum of all the resistances, which is n * r.

    Therefore, the combination would have a resistance equal to n * r, which can be written as nR.

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  • 5. 

    The current in the arm CD of the circuit will be

    • A.

      I1 + i2

    • B.

      I2 + I3

    • C.

      I1 + i3

    • D.

      I1 – i2 + i3

    Correct Answer
    B. I2 + I3
    Explanation
    The current in the arm CD of the circuit will be the sum of i2 and I3.

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  • 6. 

    The bulbs A, B and C are connected as shown in Figure. The bulbs B and C are identical. If the bulb C is fused,

    • A.

      Both A and B will glow more brighty,

    • B.

      Both A and B will glow less brightly,

    • C.

      A will glow less brightly and B will glow more brightly,

    • D.

      A will glow more brightly and B will glow less brightly.

    Correct Answer
    C. A will glow less brightly and B will glow more brightly,
    Explanation
    When bulb C is fused, it creates an open circuit in the circuit. This means that the current can no longer flow through bulb C. As a result, the current that was previously flowing through bulb C will now flow through bulb A and bulb B. Since bulb A and bulb B are connected in parallel, the total current flowing through the circuit will increase. This increase in current will cause bulb B to glow more brightly, while bulb A will receive less current and therefore glow less brightly.

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  • 7. 

    If R1 and R2 are the filament resistances of 200 watt and a 100 watt bulb respectively, both designed to run at the same voltage, then

    • A.

      R1 is four times R2

    • B.

      R2 is four times R1

    • C.

      R1 is two times 3R2

    • D.

      R2 is two times R1

    Correct Answer
    D. R2 is two times R1
    Explanation
    The answer is R2 is two times R1 because the power dissipated by a resistor is directly proportional to the square of its resistance. Since R2 is a 100 watt bulb and R1 is a 200 watt bulb, the power dissipated by R1 is twice that of R2. Therefore, R2 must have half the resistance of R1.

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  • 8. 

    If a high power heater is connected to electric mains, then the bulbs in the house become dim, because there is a

    • A.

      Current drop

    • B.

      Potential drop

    • C.

      No current drop

    • D.

      No potential drop

    Correct Answer
    A. Current drop
    Explanation
    When a high power heater is connected to the electric mains, it draws a large amount of current from the mains. This causes a drop in the current available for other devices connected to the mains, such as the bulbs in the house. As a result, the bulbs receive less current than they require, causing them to become dim. Therefore, the correct answer is "Current drop."

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  • 9. 

    A house wiring supplied with a 220 V supply line is protected by a 6 ampere fuse. The maximum number of 60 W bulbs is parallel that can be turned on is

    • A.

      11

    • B.

      22

    • C.

      33

    • D.

      66

    Correct Answer
    B. 22
    Explanation
    The maximum number of 60W bulbs that can be turned on in parallel is 22. This is because the total power consumed by the bulbs should not exceed the maximum power that can be supplied by the fuse. In this case, the maximum power that can be supplied by the fuse is 220V x 6A = 1320W. Since each bulb consumes 60W, the maximum number of bulbs that can be turned on is 1320W / 60W = 22 bulbs.

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  • 10. 

    An electric heating element consumes 500 W, when connected to a 100 V line. If the line voltage becomes 150V, the power consumed will be:

    • A.

      500 W

    • B.

      750 W

    • C.

      1000 W

    • D.

      1125 W

    Correct Answer
    D. 1125 W
    Explanation
    When the line voltage increases from 100V to 150V, the power consumed by the electric heating element will increase. This is because power is directly proportional to voltage. Using the formula P = V^2/R, where P is power, V is voltage, and R is resistance, we can see that if voltage increases, power will also increase. Therefore, the power consumed will be 1125W.

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  • 11. 

    A 100 W bulb B1 and two 60 W bulb B2 and B3 are connected to a 250V source as shown in the Figure. Now W1, W2 and W3 are the output powers of the bulbs B1, B2 and B3 respectively. Then

    • A.

      W1 > W2 = W3

    • B.

      W1 < W2 = W3

    • C.

      W1 < W2 < W3

    • D.

      W1 > W2 = W3

    Correct Answer
    C. W1 < W2 < W3
    Explanation
    The power output of a bulb is directly proportional to the voltage across it. Since all the bulbs are connected to the same 250V source, the bulb with the highest wattage (100W) will have the highest power output (W1), followed by the bulbs with lower wattage (60W each), resulting in W1 < W2 < W3.

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  • 12. 

    In a copper voltameter, mass deposited in thirty seconds is m grams. If the time current graph is as shown in figure. E.C.E. of copper is (current in milli ampere)

    • A.

      0.1 m

    • B.

      0.5 m

    • C.

      0.6 m

    • D.

      M

    Correct Answer
    B. 0.5 m
    Explanation
    The correct answer is 0.5 m. This can be determined by observing the time-current graph in the figure. The graph shows that the current remains constant at a certain value for the entire 30 seconds. According to Faraday's laws of electrolysis, the mass of a substance deposited during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. Since the current remains constant for the entire 30 seconds, it can be inferred that the quantity of electricity passed is also constant. Therefore, the mass deposited in 30 seconds is directly proportional to the constant quantity of electricity passed, which is represented by the value 0.5 m.

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  • 13. 

    Consider the following two statements (i) Free-electro density is different in different metals. (ii) Free-electro density in a metal depends on temperature. Seebeck effect is caused

    • A.

      Due to (i) but not due to (ii)

    • B.

      Due to (ii) but not due (i)

    • C.

      Due to both (i) and (ii)

    • D.

      Neither due to (i) nor due to (ii)

    Correct Answer
    C. Due to both (i) and (ii)
    Explanation
    The Seebeck effect is caused due to both (i) and (ii) because it relies on the presence of a temperature gradient and the difference in free-electron density in different metals. The temperature gradient creates a flow of electrons from the hot end to the cold end of a conductor, and the difference in free-electron density between different metals determines the magnitude of this flow. Therefore, both factors are necessary for the Seebeck effect to occur.

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  • 14. 

    If the electric current in a lamp decreases by 5 %, then the power output decreases by:

    • A.

      2.5%

    • B.

      5%

    • C.

      10%

    • D.

      20%

    Correct Answer
    C. 10%
    Explanation
    When the electric current in a lamp decreases by 5%, the power output decreases by 10%. This is because power is directly proportional to the square of the current. Therefore, if the current decreases by 5%, the power output will decrease by (5%)*(5%) = 25%, which is equivalent to a 10% decrease.

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  • 15. 

    The thermo e.m.f. of a thermo couple varies with the temperature of the hot junction as E = a 𝜃 + b 𝜃2 in bolts where the ratio a/b is 700°C. If the cold temperature is kept at 0°C, then the neutral temperature is

    • A.

      700°C

    • B.

      350°C

    • C.

      1400°C

    • D.

      No neutral temperature is possible for this thermocouple

    Correct Answer
    D. No neutral temperature is possible for this thermocouple
    Explanation
    The given equation E = a𝜃 + b𝜃^2 represents the relationship between the thermo e.m.f. (E) and the temperature (𝜃) of the hot junction. The ratio a/b is given as 700°C. In order to find the neutral temperature, we need to find the temperature at which the thermo e.m.f. becomes zero. However, since the equation contains a quadratic term (b𝜃^2), it will always have a non-zero value and will never reach zero. Therefore, no neutral temperature is possible for this thermocouple.

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Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

  • Current Version
  • Mar 22, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Nov 25, 2013
    Quiz Created by
    Tanmay Shankar
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