All, Please just help me in this, I will need you to please give me the clue-thanks.
a flashlamp bulb is marked '2.5V,0.30A' and has to be operated from a dry battery of e.m.f 3.0V for the current p.d. of 2.5V to be produced across it. Why?
To understand why a flashlamp bulb marked "2.5V, 0.30A" needs to be operated from a 3.0V dry battery for a current potential difference (p.d.) of 2.5V to be produced across it, we need to consider a few key concepts.
1. Voltage (V): Voltage, measured in volts (V), represents the potential difference between two points in an electric circuit. It is the driving force that allows electric current to flow.
2. Current (I): Current, measured in amperes (A), represents the flow of electric charge through a circuit. It is the rate at which electric charges move.
3. Ohm's Law: Ohm's Law states that the current (I) flowing through a conductor is directly proportional to the voltage (V) applied across it and inversely proportional to the resistance (R) of the conductor. Mathematically, this can be expressed as V = I * R.
Now, let's apply these concepts to the given scenario. The flashlamp bulb is marked "2.5V, 0.30A," which means it requires a potential difference of 2.5V to produce a current of 0.30A. However, the dry battery being used has an electromotive force (emf) of 3.0V.
Since the flashlamp bulb has a resistance (R), we can use Ohm's Law to calculate it. A resistance can be calculated using the formula R = V / I, where V is the voltage across the bulb and I is the current flowing through it.
Given that the voltage across the bulb should be 2.5V and the current through it is 0.30A, we can calculate the resistance of the bulb as follows:
R = 2.5V / 0.30A = 8.33 Ohms
However, the battery being used has an emf of 3.0V, which means it can provide a slightly higher voltage than 2.5V. This is necessary to overcome the internal resistance of the battery and maintain a steady current flow through the bulb.
By using a 3.0V battery, the flashlamp bulb will receive a potential difference of 2.5V, providing the required voltage for it to operate properly and produce the desired illumination.