Change in voltage due to additional source

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When an identical voltage source is added in parallel to the original, the effective internal resistance is halved, but the voltage across the load remains unchanged at V. In contrast, connecting an identical source in series doubles the equivalent resistance, resulting in a new voltage of 2V across the load. The discussion highlights the importance of considering the load resistance when analyzing the effects of additional voltage sources. The original question's wording is deemed unclear, particularly regarding the load's nature. Overall, the addition of sources affects current and resistance differently depending on the configuration.
Queequeg
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Homework Statement



A voltage source connected to a circuit produces a voltage V, current I and has internal resistance R. What is the new voltage if an identical voltage source is added to the original:

a) in parallel
b) in series

Homework Equations



V=IR[/B]

The Attempt at a Solution



a. For a voltage source in parallel, the equivalent resistance is R/2. Each source provides a current I so the total current is I+I=2I. Therefore the new voltage is V_n=(2I)(R/2)=IR=V unchanged

b. The equivalent resistance is 2R and the current is I because they are in series, so the new voltage is V_n=I(2R)=2V doubled.
 
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Queequeg said:

Homework Statement



A voltage source connected to a circuit produces a voltage V, current I and has internal resistance R. What is the new voltage if an identical voltage source is added to the original:

a) in parallel
b) in series

Homework Equations



V=IR[/B]

The Attempt at a Solution



a. For a voltage source in parallel, the equivalent resistance is R/2. Each source provides a current I so the total current is I+I=2I. Therefore the new voltage is V_n=(2I)(R/2)=IR=V unchanged

b. The equivalent resistance is 2R and the current is I because they are in series, so the new voltage is V_n=I(2R)=2V doubled.
What is the effect of the circuit to which the voltage supplies are connected? Presumably the voltage mentioned is meant to be the voltage across that load?

Fig1.gif


When you connect an identical source in parallel with the first, I agree that the effective internal resistance is halved while the effective cell voltage remains the same. But how will that effect the voltage V that is measured across the load resistance?
 
The original question is badly worded. I suppose you had better assume the load is resistive.
 
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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