Find power dissipated in circuit with a ground

[physics416]

Homework Statement

1. Find the power dissipated by 12 ohm resistor
2. What is the potential at points a,b,c,d?

Homework Equations

P=I^2R
I=V/R

The Attempt at a Solution

For a) 3v-IR-6v-IR=0
9v=I(18)
I=0.5
P=0.5^2 * 12 = 3 watts
But not sure if the ground changes anything

b) unsure if ground changes anything

 

[Noctisdark]

Hint : What is the voltage btw point b,a and b,c and a,d ? Groud represents 0V

 

[physics416]

Between bc is 6v and between ad is 3v? if my current calculation is correct, then is v between ab is 3? But are those the potential differences or the actual potentials? For example does the 6 ohm resistor use up all its potential such that it is 0 at point b and point c gains 6v?

 

[physics416]

Hint : What is the voltage btw point b,a and b,c and a,d ? Groud represents 0V

Between bc is 6v and between ad is 3v? if my current calculation is correct, then is v between ab is 3? But are those the potential differences or the actual potentials? For example does the 6 ohm resistor use up all its potential such that it is 0 at point b and point c gains 6v?

 

[Noctisdark]

Edit : for mistakes, Vad = 3V, Va = 0 since it's connected to the ground so Vd = 3, Vbc = 6V just as you said, according to your calculation P = RI^2 = V^2/R = 3, you're almost there !,

 

[SammyS]

Homework Statement

1. Find the power dissipated by 12 ohm resistor
2. What is the potential at points a,b,c,d?

Homework Equations

P=I^2R
I=V/R

The Attempt at a Solution

For a) 3v-IR-6v-IR=0
9v=I(18)
I=0.5
P=0.5^2 * 12 = 3 watts
But not sure if the ground changes anything

b) unsure if ground changes anything

The ground changes nothing.

Your solution is correct.

 

[ehild]

b) unsure if ground changes anything

The "ground" means that the potential is zero at point a. But that does not change the potential differences between two points of the circuit.

 

[CWatters]

For a) 3v-IR-6v-IR=0

You got the right answer but this equation is inconsistent.

+3v implies you are summing the voltages anticlockwise so to be consistent it should be +6v not -6V...

+3V - (I*12) +6V - (I*6) = 0