Power in 6.0 Ohm Resistor - Circuit Related Q

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To determine the power dissipated in a 6.0-ohm resistor, start by calculating the currents in the circuit using Ohm's Law (V = IR) and Kirchhoff's laws. The power can be calculated using the formula P = I^2R after finding the current through the resistor. The total current in the circuit is approximately 2.46 A, but individual currents for each resistor need to be determined for accurate power calculations. Understanding Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL) will aid in solving for these individual currents. This approach is essential for analyzing complex circuits with multiple components.
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Determine the power dissipated in the 6.0 resistor in the circuit shown in the drawing. (R1 = 4.0 , R2 = 6.0 and V1 = 15 V.)

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Am I supposed to be using P = V^2/R..is the power flowing through each resistor going to be diff?

Anyone want to help me out where to start?
 
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You can solve this many different ways. First note that the current through the 2ohm and the 1ohm resistor is the same, and that the current through R2 and the 1ohm resistor is the same.

I would first solve for the currents, and then you know that:

P=IV

V=IR

Thus,

P=I(IR)=I^2R

Do you know KVL, KCL?
 
How exactly can I get the currents for each individual peice?

I know that the total current is going to be V = IR..which equals 2.46..

But how can I solve for individual currents? Since you don't have V..what can you use to get the I?
 
Last edited:
MD2000 said:
How exactly can I get the currents for each individual peice?

I know that the total current is going to be V = IR..which equals 2.46..

But how can I solve for individual currents? Since you don't have V..what can you use to get the I?
Think Kirchoff's laws, you can use these the calculate the voltage or current at each point in the circuit. :wink:
 
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