The hot resistance of a flashlight bulb is , and it is run by a alkaline cell having a internal resistance. (a) What current flows? (b) Calculate the power supplied to the bulb using . (c) Is this power the same as calculated using?
(a)The current flows 0.658 A .
(b)The power supplied is 0.997 W .
(c)Yes, the result is the same.
Alkaline cell is use as the power source for various proposes. The emf of alkaline cell is given by equation,
Here, r is the internal resistance of the battery, I is the current through the battery, and is the resistance of the bulb.
The flashlight, whose bulb has a hot resistance and is run by a alkaline cell with internal resistance .
The current that flows through the circuit is obtained using the loop rule as
Therefore, the current flows 0.658 A .
The power supplied to the bulb is
Therefore, the power supplied is 0.997 W .
Calculation of the power supplied to the bulb using the formula if for V can be use the terminal voltage of the cell, because this is the same as the voltage of the light-bulb . The terminal voltage is
The power is then
Therefore, the result is the same.
The duration of a photographic flash is related to an RC time constant, which is 0.100 µs for a certain camera. (a) If the resistance of the flash lamp is 0.0400 Ω during discharge, what is the size of the capacitor supplying its energy? (b) What is the time constant for charging the capacitor, if the charging resistance is 800 kΩ?
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