Change in voltage due to additional source

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 1K views
Queequeg
Messages
24
Reaction score
0

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.
 
Physics news on Phys.org
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?