Question

A 30V-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

• A. $10\Omega$
• B. $30\Omega$
• C. $20\Omega$
• D. $40\Omega$

Answer 1

Answer: (B)

$30\Omega$

Resistance of lamp
$R_0=\frac{V^2}{P}=\frac{(30{)}^2}{90}=10\Omega$
Current in the lamp
$I=\frac{V}{R_0}=\frac{30}{10}=3A$
As the lamp is operated on 120V DC, then resistance becomes
$R^{\prime }=\frac{V^{\prime }}{i}=\frac{120}{3}=40\Omega$
For proper glow, a resistance R is joined in series with the bul
$R^{\prime }=R+R_0$
$\Rightarrow R^{\alpha }=R^{\prime }-R_0=40-10=30\Omega$