# Finding the current dissipated on each resitor

I'm given a question and part (1 asks me to find the total resistance of a 3.0 and 6.0 ohm resistor when I) in series and ii) In parallel. The answers to those are 9 ohms and 2 ohms respectively.

Now, part B) says that a 4.0 ohm and 6.0v battery of negligible internal resistance are connected in series with the combination In (a)

Now it wants me to find the current, p.d., and power dissipated on each resitor but I'm not sure how. I tried several attempt but they've not been successful.

Thanks.

• the current in a series circuit is the same at all points in that circuit which applies to the diagram shown. Nov 6 '17 at 9:15
• Intersting in that some of us provide hints to get the OP to improve, others just shove out the answer.... Nov 6 '17 at 9:17
• "a 6.0 v battery and 4.0 ohm resistor of negligible internal resistance" doesn't make sense. The part "of negligible internal resistance" must belong to battery. A resistor of negligible internal resistance is nonsence.
– Curd
Nov 6 '17 at 10:03
• Yes sorry it was the other way @Curd Nov 6 '17 at 10:13

The only voltage source in the system is the 6V battery, so use $\frac{V}{R_1+R_2+R_3}$ to find the current $I$ going through each resistor.
Now $V=IR$ and $P = IV$. You can substitute to get $P = I^2 × R$. You can now find the power dissipated by each resistor.