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# Electricity(circuits) watch

1. https://00c07169b8f01b5b433281e611ee...%20Physics.pdf

Question 6

if the pd across the 400 ohm resistors is 20V then is the voltage across one of them 10V then?

also help with 6b part 2 i don't understand why you work out the current flowing through one of the 400 ohm resistors and then you multiply by 2 to get the current going through the 25 ohm resistor
2. (Original post by thefatone)
samb1234
Can you post a picture of the question? Nit on my laptop atm
3. (Original post by thefatone)
https://00c07169b8f01b5b433281e611ee...%20Physics.pdf

Question 6

if the pd across the 400 ohm resistors is 20V then is the voltage across one of them 10V then?

also help with 6b part 2 i don't understand why you work out the current flowing through one of the 400 ohm resistors and then you multiply by 2 to get the current going through the 25 ohm resistor
For 6(a) Use rules for resistance in parallel and in series. you haven'y asked about this so I assume you are comfortable with these.

6(b)i - you will hopefully know that

For 6(b)ii You need to use and Kirchoff's first law.
4. (Original post by thefatone)
https://00c07169b8f01b5b433281e611ee...%20Physics.pdf

Question 6

if the pd across the 400 ohm resistors is 20V then is the voltage across one of them 10V then?

also help with 6b part 2 i don't understand why you work out the current flowing through one of the 400 ohm resistors and then you multiply by 2 to get the current going through the 25 ohm resistor
First part is just using the rules for adding resistance in parallel and series. In series, Rt=R1 +R2... , in parallel 1/Rt =1/R1 +1/R2... . Then for the next part if we know the power dissipated, and we know that p=V^2/R so we can easily work out V. For the current one, kirchoffs current laws tell us that the current will split at the junction so the current through the 25ohm resistor will be twice that of the 400 ohm ones
5. (Original post by samb1234)
First part is just using the rules for adding resistance in parallel and series. In series, Rt=R1 +R2... , in parallel 1/Rt =1/R1 +1/R2... . Then for the next part if we know the power dissipated, and we know that p=V^2/R so we can easily work out V. For the current one, kirchoffs current laws tell us that the current will split at the junction so the current through the 25ohm resistor will be twice that of the 400 ohm ones
i see thanks

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