Potential Difference: Effects of Voltmeter Connection

AI Thread Summary
Connecting a voltmeter across a source of EMF, such as a battery, can affect the measured potential difference due to the voltmeter's finite resistance. While ideal voltmeters should have infinite resistance to prevent current alteration, real voltmeters do influence the current flow. As current flows through the battery, internal resistance leads to a decrease in the potential difference measured by the voltmeter. However, the electromotive force (EMF) of the source remains unchanged, as it represents the energy supplied per unit charge when no current is flowing. Thus, the voltmeter's connection alters the observed potential difference but not the actual EMF of the source.
tin llenaresas
Messages
18
Reaction score
0
what happens to the potential difference of a source of emf when a voltmeter is connected across its terminals?
tanx. :smile:
 
Physics news on Phys.org
HINT: Does current flow when the voltmeter is connected?
 
hey tide,

ideal voltmeters should have infinite resistance so as not to alter current but real voltmeters have finite resistance.does our voltmeter, in any way, still alter the current flowing through our set-up?
 
I assume that the source of EMF that you have in mind is a battery. Batteries have internal resistance and if there is current flowing through the battery then, by Ohm's Law, there must be a potential difference across the resistance.
 
so, does the voltmeter affects the potential difference of a source?
 
Yes, the potential difference across the battery will decrease but still the emf of the source remains the same. The emf of a source is the energy supplied by the source per unit charge and is measured by the potential difference across the terminals, when no current is flowing through the source.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top