When the switch is closed, it provides a path with no resistance so you could see it as short circuiting the resistor.
Hope this gave a bit of help with giving away too much,
Still ask if it you're still unsure :P
Hope this gave a bit of help with giving away too much,
Still ask if it you're still unsure :P
Original post by JustMatcha
When the switch is closed, it provides a path with no resistance so you could see it as short circuiting the resistor.
Hope this gave a bit of help with giving away too much,
Still ask if it you're still unsure :P
Hope this gave a bit of help with giving away too much,
Still ask if it you're still unsure :P
is it something to do with it being an ohmic component so as the current is allowed to flow the temperature of the bulb increases and the resistance increases? and the current flowing through resistor is directly proportional to the voltage across it
i don’t know i’m just guessing and rlly unsure x
Original post by jlcommander
is it something to do with it being an ohmic component so as the current is allowed to flow the temperature of the bulb increases and the resistance increases? and the current flowing through resistor is directly proportional to the voltage across it
i don’t know i’m just guessing and rlly unsure x
i don’t know i’m just guessing and rlly unsure x
Btw I can still be wrong but I feel you're overcomplicating it xd
So when the switch is closed, it gives the charges two routes: through a resistor which requires work or through just a wire
Therefore, the current completely ignores the resistor route and flows through the switch so you can just picture it as a circuit diagram with the battery and the lamp.
Since there's no current through the resistor it has no potential difference.
In turn, the lamp will have all the battery's potential difference because it no longer has to share with the resistor!
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