A $30\,V-90\,W$ 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
- 10 $\Omega$
- 30 $\Omega$
- 20 $\Omega$
- 40 $\Omega$
The Correct Option is B
Solution and Explanation
$R_{0} = \frac{V^{2}}{P} = \frac{\left(30\right)^{2}}{90} = 10\,\Omega $
The current in the lamp will be
$I = \frac{V}{R_{0} } = \frac{30}{10} = 3A $
As the lamp is operated on 120V DC, then resistance becomes
$ R' = \frac{V'}{i} = \frac{120}{3} = 40 \Omega$
For proper glow, a resistance R is put in series with the bulb so that
$ R' = R + R_{0} $
$ \Rightarrow R^{\alpha} = R' - R_{0} = 40 - 10 = 30 \,\Omega$
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Concepts Used:
Power
Power is the rate of doing an activity or work in the minimum possible time. It is the amount of energy transferred or converted per unit of time where large power means a large amount of work or energy.
For example, when a powerful car accelerates speedily, it does a large amount of work which means it exhausts large amounts of fuel in a short time.
The formula of Power:
Power is defined as the rate at which work is done upon an object. Power is a time-based quantity. Which is related to how fast a job is done. The formula for power is mentioned below.
Power = Work / time
P = W / t
Unit of Power:
As power doesn’t have any direction, it is a scalar quantity. The SI unit of power is Joules per Second (J/s), which is termed as Watt. Watt can be defined as the power needed to do one joule of work in one second.