Join TSR now and get all your revision questions answeredSign up now

Help with a 1 mark as level electricity question? Watch

    • Thread Starter
    Offline

    2
    ReputationRep:
    Name:  Capture.PNG
Views: 21
Size:  14.3 KBPlease post a method.
    Offline

    3
    ReputationRep:
    (Original post by Jas1947)
    Name:  Capture.PNG
Views: 21
Size:  14.3 KBPlease post a method.
    You need to break it down into simpler chunks.

    we know the PD across RA and RB is the same because they are in parallel with each other.
    and the resistance of RB is 2x the resistance of RA so the current in RB must be some proportion of the current in RA because I=V/R
    and we know power is given by P=IV
    so we know the power dissipation of RB as a proportion of the power dissipated in RA

    -----
    Now the current in RC will be equal to the sum of the currents in RA and RB
    so we can find the power dissipation in RC using the rule P=I2R

    then we just need to add the 3 power dissipations together to get the total being supplied by the battery.
 
 
 
Poll
How are you feeling about Results Day?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.