Bloody hell! I recall struggling with this myself 35 years ago (!) and here I am, trying to help DS1! He's doing higher tier fast-paced triple science- and shouldn't be. We now know that. His school only offer triple at 'fast pace' and we were assured DS was up to it, but he isn't, and we're stuck with it!
So here we are, me trying to come to terms with what happens when you stick resistance into a circuit in series with one in parallel. Or something.
He, no we 
have to find the following- wish I could do circuit diagrams on MN!: We have a circuit diagram presenting both 'in series' and 'in parallel' components. We're asked ? the voltage of a certain battery, the ammeter immediately beyond it is also '?'; beyond that the circuit splits! The first split has another ? ammeter, then a bulb of unknown (but not required resistance- all the 4 bulbs have the same resistance in this question)- we're asked ? the voltage across this bulb. This split then rejoins the whole circuit back to the cell. The second split starts with an ammeter measuring 1A- beyond this we have 3 identical bulbs (each having the same unknown resistance as the bulb on the first split); We're asked for the voltage across each of these bulbs though we're told the first has 3V across it. This second split then also rejoins the original circuit back to the cell.
Jeez.
My ishooz and suggestions:
-
If all the bulbs present the same resistance, surely as we're given the voltage across one of them i.e. 3V, even though 3 are in series and one is in parallel with them, they all have a voltage of 3V across them? So it's a 3V battery?
- If 1 amp (as we're told) services a series of 3 bulbs and if 'bigger resistance = lower current' the ratio is "one amp to the 3 resistors, therefore 3 amps to the one resistor" (the one in parallel to these three) I can work out that the first split's ammeter should read 3A.
AM I right?
And congratulations if you've stuck with me!!