Physics Electricity Inquiry
This is electricity theory, just dont understand this rule, ill attatch the ciruit diagram and question.
My understanding is that, algebraically, if the one of the components, in parallel, were to stop conducting.
This would result in a higher resistance, decreasing the flow of current, and causing the overall voltage to decrease.
But, V=IR so wouldn't an increase in resistance, increase the PD between two points in a circuit
Which leads onto part ii)
Lamp Y has increased resistance, and therefore PD will increase. Therefore a greater rate of power, via P=IV
I'm just kinda unsure between i and ii)
It seems contradictory in a sense, the theory makes sense, but the relationship contradicts it and I am just overcomplicating it.
I've attatched the file
My understanding is that, algebraically, if the one of the components, in parallel, were to stop conducting.
This would result in a higher resistance, decreasing the flow of current, and causing the overall voltage to decrease.
But, V=IR so wouldn't an increase in resistance, increase the PD between two points in a circuit
Which leads onto part ii)
Lamp Y has increased resistance, and therefore PD will increase. Therefore a greater rate of power, via P=IV
I'm just kinda unsure between i and ii)
It seems contradictory in a sense, the theory makes sense, but the relationship contradicts it and I am just overcomplicating it.
I've attatched the file
(edited 1 year ago)
R 1 on the left and the parallell circuit on the form a potential divider.
When the resistance on the right increases (as you correctly state) the pd increases on the right.
The total pd across left and right has to continue to add up to 24V so the PD on the left must go down.
Overall current can go down at the same time as current through the remaining bulb Y goes up so overall current doesn't help you explain the brightness of the remaining bulb.
eg the bulbs could be passing one amp each before the bulb failure which is an overall current of 2 amps
After the bulb failure the remaining bulb could now be passing 1.2 amps... overall current is down to 1.2A but the current through the remaining bulb has gone up from 1 to 1.2A
When the resistance on the right increases (as you correctly state) the pd increases on the right.
The total pd across left and right has to continue to add up to 24V so the PD on the left must go down.
Overall current can go down at the same time as current through the remaining bulb Y goes up so overall current doesn't help you explain the brightness of the remaining bulb.
eg the bulbs could be passing one amp each before the bulb failure which is an overall current of 2 amps
After the bulb failure the remaining bulb could now be passing 1.2 amps... overall current is down to 1.2A but the current through the remaining bulb has gone up from 1 to 1.2A
Original post by Joinedup
R 1 on the left and the parallell circuit on the form a potential divider.
When the resistance on the right increases (as you correctly state) the pd increases on the right.
The total pd across left and right has to continue to add up to 24V so the PD on the left must go down.
Overall current can go down at the same time as current through the remaining bulb Y goes up so overall current doesn't help you explain the brightness of the remaining bulb.
eg the bulbs could be passing one amp each before the bulb failure which is an overall current of 2 amps
After the bulb failure the remaining bulb could now be passing 1.2 amps... overall current is down to 1.2A but the current through the remaining bulb has gone up from 1 to 1.2A
When the resistance on the right increases (as you correctly state) the pd increases on the right.
The total pd across left and right has to continue to add up to 24V so the PD on the left must go down.
Overall current can go down at the same time as current through the remaining bulb Y goes up so overall current doesn't help you explain the brightness of the remaining bulb.
eg the bulbs could be passing one amp each before the bulb failure which is an overall current of 2 amps
After the bulb failure the remaining bulb could now be passing 1.2 amps... overall current is down to 1.2A but the current through the remaining bulb has gone up from 1 to 1.2A
Thanks for the explanation, so theoretically is now a series circuit where the voltage is now shared. Due to conversation, the PD on R1 therefore must decrease
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