The current passing through the bulb is directly related to the voltage applied across it and the resistance of the bulb. Using Ohm's Law (I = V/R), we can calculate the current flowing through the bulb by knowing the voltage and resistance values. Additionally, the brightness of the bulb can also be an indicator of the current passing through it, as higher current typically results in a brighter bulb.
When the two electrodes contact, it completes the circuit, allowing the current to flow from one electrode to the other through the bulb. The current passing through the filament of the bulb causes it to heat up and emit light, resulting in the bulb lighting up.
Added resistance in a circuit will decrease the current flowing through the circuit, resulting in a decrease in the brightness of the bulb. This is because the bulb's brightness is directly proportional to the current passing through it. More resistance means less current, which leads to reduced brightness.
To determine the number of electrons passing through the bulb in one minute, we need to use the formula: Number of electrons = (current * time)/(charge of one electron) Given current = 300 A and time = 1 minute, we can calculate the number of electrons passing through the bulb. Charge of one electron is approximately 1.6 x 10^-19 Coulombs.
The bulb that glows brightly likely has a higher current passing through it compared to the dim bulb. This could be due to differences in resistance, voltage, or power ratings of the two bulbs.
An increase in the current through a bulb will increase its light output because more current means more electrons passing through the filament, which generates more heat and light. Conversely, a decrease in current will result in lower light output as there are fewer electrons flowing through the filament to produce light.
We can infer that current is needed to generate Electrical appliances and also to make the flow of electric circuits be completed.
what can be inferred about the voltage across the bulb
This is usually done by passing an electrical current through the wire.
Yes but very slightly,because temperature coefficient of bulb element is very low.
When the two electrodes contact, it completes the circuit, allowing the current to flow from one electrode to the other through the bulb. The current passing through the filament of the bulb causes it to heat up and emit light, resulting in the bulb lighting up.
Added resistance in a circuit will decrease the current flowing through the circuit, resulting in a decrease in the brightness of the bulb. This is because the bulb's brightness is directly proportional to the current passing through it. More resistance means less current, which leads to reduced brightness.
The effect of current utilised in a bulb is the conversion of electrical energy into light energy through the heating of the filament inside the bulb. This process is known as resistive heating, where the current passing through the filament encounters resistance, causing it to heat up and produce light.
To determine the number of electrons passing through the bulb in one minute, we need to use the formula: Number of electrons = (current * time)/(charge of one electron) Given current = 300 A and time = 1 minute, we can calculate the number of electrons passing through the bulb. Charge of one electron is approximately 1.6 x 10^-19 Coulombs.
Your standard light bulb where a filament is heated by a current passing through it. The heated filament then gives off light.
The bulb that glows brightly likely has a higher current passing through it compared to the dim bulb. This could be due to differences in resistance, voltage, or power ratings of the two bulbs.
An increase in the current through a bulb will increase its light output because more current means more electrons passing through the filament, which generates more heat and light. Conversely, a decrease in current will result in lower light output as there are fewer electrons flowing through the filament to produce light.
The current flowing through a bulb is equal to the (voltage across the bulb) divided by the (bulb resistance), and can be expressed in Amperes. The rate at which the bulb dissipates energy is equal to (voltage across the bulb) times (current through the bulb), and can be expressed in watts.