Calculating Power Dissipation in a Wired Circuit with Resistors and a 9V Battery

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The discussion focuses on calculating the power dissipation in a 6-ohm resistor (R4) within a circuit powered by a 9V battery. The total resistance of the circuit is determined to be 5.1 ohms, resulting in a total current of 1.76A. The voltage across the combined resistors R4 and R5 is calculated to be 3.7V. To find the current through R4 specifically, the voltage across the combined resistors is divided by their total resistance, allowing for the correct power dissipation calculation using the formula P = I^2R. The final calculations confirm the correct approach and yield the desired result for power dissipation in R4.
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ooooooXXXXR2XXXXXR3XXXX
XXR1XXoooooooooooooooooXXXXX
XooooXoooooooooooooooooXooooX
XooooXXXXR4XXXXXXR5XXXXoooX
XoooooooooooooooooooooooooooX
XoooooooooooooooooooooooooooX
XoooooooooooooooooooooooooooX
XXXXXXXXX+BATTERY-XXXXXXXXX

X=WIRE
o= blank space (ignore)
R=resistor
battery= 9 V

Determine the power dissipated in the R4 resistor in the circuit shown in the drawing.

R1=3 ohms
R2=2 ohms
R3=1 ohms
R4=6 ohms
R5=1 ohms

So here is what i know so far. The resistance of R4+R5=7. The resistance of R2+R3=3. So the resistance of R2345=1/(1/7+1/3)=2.1 ohms. The total resistance through the entire circuit is R1+R2345=3+2.1=5.1 ohms.
So I use this total resistance to calculate the total current I=V/R so
I=9V/5.1ohms=1.76A. Now I can use this total current to calculate the voltage across R4 and R5. V=IR=1.76(2.1)=3.7 V across R4 and R5 since Voltage across parallel circuits are the same. I know power=IR^2=V^2/R.
But i cannot just plug in 3.7V^2/6ohms. This isn't giving me the right answer. I need to find the voltage across R4 only. I am not sure how to do this since the current is different passing through each resistor. please help and correct me if any of the previous steps i did were wrong. thanks.
 
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spoonthrower said:
ooooooXXXXR2XXXXXR3XXXX
XXR1XXoooooooooooooooooXXXXX
XooooXoooooooooooooooooXooooX
XooooXXXXR4XXXXXXR5XXXXoooX
XoooooooooooooooooooooooooooX
XoooooooooooooooooooooooooooX
XoooooooooooooooooooooooooooX
XXXXXXXXX+BATTERY-XXXXXXXXX

X=WIRE
o= blank space (ignore)
R=resistor
battery= 9 V

Determine the power dissipated in the R4 resistor in the circuit shown in the drawing.

R1=3 ohms
R2=2 ohms
R3=1 ohms
R4=6 ohms
R5=1 ohms

So here is what i know so far. The resistance of R2+R3=7. The resistance of R4+R5=3. So the resistance of R2345=1/(1/7+1/3)=2.1 ohms. The total resistance through the entire circuit is R1+R2345=3+2.1=5.1 ohms.
So I use this total resistance to calculate the total current I=V/R so
I=9V/5.1ohms=1.76A. Now I can use this total current to calculate the voltage across R4 and R5. V=IR=1.76(2.1)=3.7 V across R4 and R5 since Voltage across parallel circuits are the same. I know power=IR^2=V^2/R.
But i cannot just plug in 3.7V^2/6ohms.
I have no idea where your 6 ohms comes from!
This isn't giving me the right answer. I need to find the voltage across R4 only. I am not sure how to do this since the current is different passing through each resistor. please help and correct me if any of the previous steps i did were wrong. thanks.
You kno wthe voltage across the combined R45. Now find the current through both by using I = V/R45 = 3.7V/3 ohms. That's the current through R4. Now use Power = R_4 I^2
 
sry, lil typo, it is fixed now.
R4=6 ohms
and I got the right answer. thanks man.
 
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