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A 30 V, 90 W lamp is to be operated on a 120 V D. C. line. For proper glow, a resistor of ____ ohm should be connected in series with the lamp.
1. 40
2. 10
3. 20
4. 30

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Correct Answer - Option 4 : 30


  • Electric Power: The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

\(P = \frac{W}{t} = VI = {I^2}R = \frac{{{V^2}}}{R}\)

Where V = Potential difference, R = Resistance and I = current.


Given - Voltage rating (V) = 30 V, Power rating of the lamp (P) = 90 W

  • The resistance of the lamp can be calculated as

\(⇒ R_L= \frac{{{V^2}}}{P}=\frac{(30)^2}{90}=10\, Ω\)

Now, the current in the lamp is 

\(⇒ I_L=\frac{V}{R_L}=\frac{30}{10}=3\, A\)

  • According to the question, as the lamp is operated on 120V DC, then resistance becomes

\(⇒ R'=\frac{V'}{I_L}=\frac{120}{3}=40\, Ω\)

  • For proper glow, a resistor must be connected in series with the lamp, therefore 

⇒ R' = R + RL

⇒ R = R' - RL = (40 - 10) Ω = 30 Ω

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