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1. Hi I need some help with this physics question

From the circuit in the figure calculate the following:

a) Currents through resistors R1, R2 and R3
b) Power dissipated in R3

Note: values of the resistances are in Ohms.

Attached Images

2. For A:
R1, R3: 10 = I(3 + 7)
I = 1A

R2, R3: 5 = I(3 + 14)
I = 0.29A
Just one more step to work out the current of each resistor.

For B:
Use v = IR with the current from A.
3. hey! what do you mean one more step for question a.. dont get that part
4. (Original post by SpikeSpiegel)
hey! what do you mean one more step for question a.. dont get that part
You have IR1 and IR2.
IR3 = IR2 + IR1.
5. Cool ... thank you so much!!!!
6. For part a) i got

a) R1, R3 : 10 = I(3+7)
Therefore I = 1A

R2, R3 : 5 = I(3+14)
I= 0.29 A

I have - IR1 and IR2
: IR3 = IR2 + IR1

So, R3 = R1+R2
R3= 1A + 0.29A = 1.29A

is this how I would approach this?

I don't know how to go about part b since there is 2 voltages for R3
7. (Original post by SpikeSpiegel)
For part a) i got

a) R1, R3 : 10 = I(3+7)
Therefore I = 1A

R2, R3 : 5 = I(3+14)
I= 0.29 A

I have - IR1 and IR2
: IR3 = IR2 + IR1

So, R3 = R1+R2
R3= 1A + 0.29A = 1.29A

is this how I would approach this?

I don't know how to go about part b since there is 2 voltages for R3
Yes.

For b, use v = IR, you know I and R.
8. but it asks for power dissipated ... not voltage?
9. (Original post by SpikeSpiegel)
but it asks for power dissipated ... not voltage?

Sorry I didn't read that part of the question properly, but the idea is the same. There are two equivalent methods:
1. Calculate voltage as I said and use p = Iv.
2. Use p = I2R.
I would go with 2.
10. so.. this is how i did it

p= I^2 R

p= 1.29 x 3

so, p = 3.87?
11. (Original post by SpikeSpiegel)
so.. this is how i did it

p= I^2 R

p= 1.29 x 3

so, p = 3.87?
You forgot to square the I
It should be p = 1.292x3=4.9923W

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