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5 questions
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  • Q1
    Identify the variables in the situation above.
    Question Image
    P=60W, I=36A
    I=60A, V=36V
    V=60V, P=36W
    P=60W, V=36V
    120s
    112.39.c.5.f
  • Q2
    What equation can be used to calculate the current in the light bulb with the given information?
    Question Image
    V=IR
    P=IV
    Rp=(1/R1+1/R2+1/Rn)^-1
    Rs=R1+R2+Rn
    120s
    112.39.c.5.f
  • Q3
    What is the current moving through the bulb?
    Question Image
    1.7A
    0.6A
    2.2A
    1A
    120s
    112.39.c.5.f
  • Q4
    If the power was halved, how would the current be affected?
    Question Image
    Double
    Fourth
    Quadruple
    Half
    120s
    112.39.c.5.f
  • Q5
    Assuming that the battery providing voltage is the same voltage, what would happen to current and power if an extra resistor was added in series?
    Question Image
    No change
    Increase
    Decrease
    Not enough information
    120s
    112.39.c.5.f

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