How Do You Calculate Terminal Voltage Vab in a Circuit?

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To calculate the terminal voltage Vab in the circuit, the formula V = E - IR is applied, where E is the battery voltage and I is the current through the internal resistance. The current through the 12V battery is measured at 70.6 mA, leading to a voltage drop of 0.706V across the 10-ohm resistor. The calculation shows that Vab equals 18V after considering the current and resistances involved. The 30-ohm resistor is not directly addressed in the initial calculations, but it can be combined with the 20-ohm resistor since they are in parallel. The discussion emphasizes the importance of correctly applying circuit rules and understanding resistor configurations.
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Homework Statement


In the circuit shown in the figure below the current through the 12.0V battery is measured to be 70.6 mA in the direction shown. What is the terminal voltage Vab



Homework Equations


V=E-IR
I1=I2+I3


The Attempt at a Solution


First I set Vab= E-IR= 24 - 10(I1)
I1=I2 + I3
So I applied Kirchoff's rule on the outer loop to find I3
-12 - 10I2 + 20I3 = 0
I3= [12 + 10I2]/20 = 0.6000353 = 0.6A
I1= I2 + I3 = 70.6 mA + 0.6A = 0.6000706A = 0.6A
Vab= 24 -10I1 = 24 - 6 = 18V
Is this right?
 

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What happened to the 30-ohm resistor?

The voltage due to the internal resistance of the 12-volt battery is (70.6 mA)(10 ohms) = 706 mV = 0.706 V. You seem to have divided by another factor of 1000.
 
Can I combine the 20(ohm) with the 10(ohm) resistor?
 
No, because they're neither in series nor in parallel. You could, however, combine the 20-ohm and 30-ohm resistors because they're in parallel.
 
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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