Discussion Forum

A 30 V-90W  lamp is operated on a 120 V DC  line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance  is 

Answer:

Explanation:

Resistance of lamb

 $R_{0}= \frac{V^{2}}{P}=\frac{(30)^{2}}{90}=10$ Ω

 Current in lamp

 $I= \frac{V}{R_{0}}=\frac{30}{10}=3 A$

 As the lamp is operated on 120  V DC  , then  resistance becomes

$R'= \frac{V'}{i}=\frac{120}{3}=40$  Ω

 For proper glow , a resistance R is joined in series with the bulb

 R'= R+R₀

 $\Rightarrow  R^{\alpha}=R'- R_{0}=40-10=30$ Ω