# physics help!

#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.

0
7 years ago
#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.
0
#3
hey! what do you mean one more step for question a.. dont get that part
0
7 years ago
#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.
0
#5
Cool ... thank you so much!!!!
0
#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
0
7 years ago
#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.
0
#8
but it asks for power dissipated ... not voltage?
0
7 years ago
#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.
0
#10
so.. this is how i did it

p= I^2 R

p= 1.29 x 3

so, p = 3.87?
0
7 years ago
#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
0
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