An Electric Bulb Connected to a 220 V Power Supply Draws Current of 0 .5 A. Calculate the Power of the Bulb.

Electric power can be expressed as the rate at which work is done or energy is converted into an electric circuit. In simple words, it is the measurement of energy that is used in an interval of time.

Step 1: Formula Used

The formula of Electric Power can be expressed as,

P = VI

Here,

P = Power
V = Potential Difference
I = Electric Current

Power is also denoted as:

⇒ P = I²R
⇒ P = V²/R
The expressions above are acquired with the help of Ohm’s law, where voltage, current, and resistance are interrelated by:

V = IR

Where,

R = Resistance
V = Potential Difference
I = Electric Current

Step 2: Solution

As is given,

V = 220V
I = 0.5A

Thus, in order to find the solution, we need to determine the bulb’s power

Hence, P = VI

⇒ P = 220 x 0.5 A
⇒ P = 110 watt

Thus, the bulb’s power is 110 watts.

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An Electric Bulb Connected to a 220 V Power Supply Draws Current of 0 .5 A. Calculate the Power of the Bulb.

Step 1: Formula Used

Step 2: Solution

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An Electric Bulb Connected to a 220 V Power Supply Draws Current of 0 .5 A. Calculate the Power of the Bulb.

Step 1: Formula Used

Step 2: Solution

CBSE CLASS XII Related Questions

2.Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

3.A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

5.In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

6.Find the ratio $\dfrac{\lambda_\alpha}{\lambda_p}$ of the de Broglie wavelength $\lambda_\alpha$ associated with an alpha particle to the de Broglie wavelength $\lambda_p$ associated with a proton if both are moving with the same velocity.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

3.
A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of $60^\circ$ in 0.2 s. The value of emf induced in the loop will be:

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