a) 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?

b) ow much heat and light energy is produced by a 100W electric lamp in 5 minutes?

c) what is the resistance of a 240V,60W bulb? - In this case I got 960 ohm (am I correct?)

Yes, on c.

I will be happy to critique your work on a, b.

Please stop posting under multiple names. My patience is running thin

The bulb is rated 10w~220v it means that it can convert 10j of energy into heat and light

a) The flashlamp bulb is marked '2.5V, 0.30A', which means that it requires a voltage of 2.5V to produce the desired current of 0.30A. When operated from a dry battery with an electromotive force (emf) of 3.0V, the current p.d. (potential difference) across the bulb will be 2.5V, as indicated on the bulb. This occurs because a dry battery has an internal resistance that restricts the flow of current, resulting in a voltage drop across the battery itself. In this case, the voltage drop across the internal resistance of the battery is 0.5V (3.0V - 2.5V), which leaves the remaining 2.5V to be applied to the flashlamp bulb, satisfying its voltage requirement.

b) To determine the amount of heat and light energy produced by a 100W electric lamp in 5 minutes, we can use the equation: Energy = Power × Time.

First, we need to convert the time to seconds since power is given in watts, which is the energy expended per second.

Given: Power = 100W, Time = 5 minutes = 5 × 60 seconds = 300 seconds

For heat energy, we use the same equation, Energy = Power × Time:

Heat energy = 100W × 300s = 30,000 joules

For light energy, we assume that the electric lamp is 100% efficient, meaning that all the electrical energy input is converted into light energy. Therefore, the light energy produced is also 30,000 joules.

c) You are correct. To determine the resistance of a 240V, 60W bulb, we can use the equation for power: Power = (Voltage)^2 / Resistance.

Given: Power = 60W, Voltage = 240V

Rearranging the formula to solve for resistance:

Resistance = (Voltage)^2 / Power = (240V)^2 / 60W = 57600V^2 / 60W = 960 ohms.

a) In order to understand why the flashlamp bulb needs to be operated from a dry battery with an e.m.f of 3.0V for a current p.d. of 2.5V to be produced across it, we need to consider the concept of internal resistance.

When a battery is connected to a circuit, it has an internal resistance represented by the symbol 'r'. This internal resistance limits the flow of current through the circuit. As a result, when the bulb is connected to the battery, there will be a voltage drop across the internal resistance of the battery, reducing the voltage available across the bulb.

In this case, the marked voltage of the bulb is 2.5V. To achieve this voltage across the bulb, the battery needs to generate a higher voltage to compensate for the voltage drop across its internal resistance. Hence, a 3.0V dry battery is needed to ensure that a current p.d. of 2.5V is produced across the flashlamp bulb.

b) To calculate the heat and light energy produced by a 100W electric lamp in 5 minutes, we can use the formulas for power and energy.

Power is given by the equation:

Power = Energy / Time

Given that the power of the electric lamp is 100W and the time is 5 minutes (or 5 * 60 = 300 seconds), we can rearrange the equation to solve for energy:

Energy = Power * Time

Energy = 100W * 300s

Energy = 30,000 J

Therefore, the electric lamp produces 30,000 Joules (J) of energy in 5 minutes.

c) To find the resistance of a 240V, 60W bulb, we can use the formula for power:

Power = (Voltage^2) / Resistance

We can rearrange the equation to solve for resistance:

Resistance = (Voltage^2) / Power

Resistance = (240V)^2 / 60W

Resistance = 57600 / 60

Resistance = 960 ohms

Therefore, your calculation of 960 ohms is correct.