How Do You Calculate Current and Power in a 3-Resistor Circuit?

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To calculate current and power in a 3-resistor circuit, it's essential to set up loop equations correctly, ensuring each component is included in at least one loop. The first loop equation was deemed correct, while the second loop was identified as incorrect due to mislabeling. It's important to label nodes and clarify loop paths to avoid confusion. Each loop must complete a circuit, and shared components can appear in multiple loops; however, the current through a resistor remains consistent across equations. Properly applying Kirchhoff's Voltage Law (KVL) requires attention to the direction of current flow to determine potential drops or rises accurately.
ReidMerrill
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Homework Statement


Find the current through and power dissipated by each resistor
20160412_181936_zps2kuodzwp.jpg


Homework Equations


V=I/R
ΣIin=∑Iout

The Attempt at a Solution


20160412_183323_zps0by4tlu9.jpg

Did I set up the loop laws correctly and where do I go from here?
 
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Your first loop equation looks fine. I can't picture what your second loop is from the equation that you wrote, but I can say that it is incorrect. Your third loop equation has an error in the last term: current I2 does not pass through the 10 Ω resistor.

Can you label the nodes in your diagram and state your loop paths?

Hint: You only need enough loops so that each component is included in at least one of the loops. In this problem you can get away with two loop equations.
 
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Since the 10Ω resistor is already in the first loop do I even need to include it in the second?
 
ReidMerrill said:
Since the 10Ω resistor is already in the first loop do I even need to include it in the second?
Every loop must complete a circuit (be a closed path) and some components will be shared by more than one loop; that's expected. You must include a term for every component in the "KVL walk" that you take around given loop in order for KVL to hold.
 
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20160412_191653_zpsd6mcpiqa.jpg

I'm not sure if the current going through the 10Ω resistor is different/needs a different name
I'm also going to see if starting from the 8V battery makes more sense to me.
 
Once you've labeled a current though a component, that's the current that is associated with it. It doesn't change if you apply KVL to different loops that happen to share that component. So ##I_3## flows through the 10 Ω resistor, in the same direction as you chose it, for each KVL equation that contains that resistor. So just be careful of the signs of the terms: if you "KVL walk" over a resistor against the flow of the current, then you will "see" a potential rise instead of a drop.

In your second equation above, you are "walking" clockwise around the loop, so when you traverse the 10 Ω resistor you are proceeding upwards, against the flow of ##I_3##. Will you see a potential drop or a potential rise as a result?
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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