Rating of a bulb is 12 W, 6V. How it can be run on a 12 V supply line?
(Calculate the value of resistance to be connected with the bulb)
Answers
Answer:
In order for the proper values to match, we have to confirm what is known.
Power source is 12v.
Lamp #1 is 12Watts
However, because lamp #1 is 6v only ,it needs comparable resistance to prevent burnout of lamp at 12v.
So total of circuit must be 24W at 12V.
In order to limit current, then the total current P = E * I, solving for I gives 2 amperes.
R = E / I from ohms law.
Since E = 6 volts across ressitor (Kirchoffs rules) , then Solving for R gives 3 ohms.
But the Resistor must be able to sustain 12W, so for good measure a 15W rated 3 ohm resistor is used.
The Current through the lamp would initially be 20A, but the 3 ohm resistnce will temper that , and reduce the inrush current down to 4 amperes upon startup.
Once the Filiament starts glowing, then the current will drop down to 2 ampers in the series R and Lamp.
SInce the current is 2 amps, then the two resistaances will total 6 ohms, 3 ohms for the ballast, 3 ohms for the lamp.
Explanation:
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Explanation:
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