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A lamp is connected in series with a cap...

A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced ?

Text Solution

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When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitative reactance (`1//omegaC)` and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.
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Knowledge Check

  • A lamp is connected in series with a capacitor A) When a de source is connected to the combination, the lamp will not glow in steady state B) When ac source is connected to the combination, the brightness of glow of lamp increases with increases of frequency of ac.

    A
    A is true B is false
    B
    A is false B is true
    C
    A & B are true
    D
    A & B are false
  • (A) : An electric lamp connected in series with a variable capacitor and A.C. source, its brightness increases with increase in capacitance. (R): Capacitive reactance decreases with increase in capacitance of capacitor.

    A
    Both 'A' and 'R' are true and 'R' is the correct explanation of 'A'.
    B
    Both 'A' and 'R' are true and 'R' is not the correct explanation of 'A'
    C
    'A' is true and 'R' is false
    D
    Both 'A' and 'R' are false
  • Two identical capacitors are connected in series. Charge on each capacitor is q_0 . A dielectric slab is now introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge in each capacitor will now be

    A
    `(2q_0)/( 1+(1)/(K))`
    B
    `(q_0)/(1 +(1)/(K))`
    C
    `(2q_0)/(1+K) `
    D
    `(q_0)/(1+K)`
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