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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityElectrical PowerEasy2 minQB

PhysicsEasynumerical

A $220 V, 50 Hz$ AC source is connected to a $25 V, 5 W$ lamp and an additional resistance $R$ in series (as shown in figure) to run the lamp at its peak brightness, then the value of $R$ (in ohm) will be

Answer:

975.00

Solution:

Resistance of the bulb can be calculated as,

$P = \frac{V ^{2}}{R _{B}} \Rightarrow R_{B} = \frac{V ^{2}}{P} = \frac{2 5 ^{2}}{5} = 125Ω$ The current through the bulb for peak brightness should be,

$i = \frac{25}{125} = \frac{1}{5} A$

Now, $i_{r m s} = \frac{1}{5} = \frac{220}{R _{B} + R}$

$RB + R = 1100$

$R = 1100 - 125 = 975$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Electrical Power

⏱ 2mℹ️ Source: QB

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