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What else can I help you with?
Why emf does not depend on resistance?
Because emf is the very source of voltage, either chemical or inductive, an can be meassured at open circuit only so, internal resistance of the supplier is not affecting it.
Why you perform open circuit characteristics of a dc shunt generator?
to find the relation between emf generated and Field current .....
Which is preferred for measuring emf of a cell avoltmeter or potentiometer?
A voltmeter is used to measure voltage. A potentiometer is used to vary the amount of resistance in a circuit - it has nothing to do with measuring.
What are the factors causing a permanent magnet generator not to generate emf?
A magnetic field will not generate an EMF if there is no motion, so the primary factor causing a permanent magnet generator to not generate EMF is if it is not turning.
Is the open circuit voltage of a solar cell measured under non-illuminating condition?
Well, you should really measure the open-circuit voltage and the short circuit current both under dark and light conditions and then compare them to fully characterize a solar cell. Measuring the open-circuit voltage means measuring the voltage across the cell when no current is flowing (i.e., with a LARGE resistance as a load on the cell). Measuring the short-circuit current means measuring the current when the voltage across the circuit is essentially zero (i.e., with a VERY SMALL resistance as a load on the cell--thus, "short-circuit" current).
Why emf of cell measurd by potentiometre is accurate?
The electromotive force (emf) of a cell measured by a potentiometer is accurate because a potentiometer measures the potential difference between the two electrodes without drawing any current from the cell, leading to minimal disturbance in the cell's internal resistance. This allows for a more precise measurement of the emf of the cell under open circuit conditions.
What is the Condition for terminal voltage across a secondary cell to be equal to its emf?
The condition for the terminal voltage across a secondary cell to be equal to its emf is when there is no current flowing through the cell. When there is no current, there is no voltage drop across the internal resistance of the cell, and thus the terminal voltage equals the emf.
How do you calculate total emf in series connecte cells?
With cells connected in series, the total emf of the 'stack' is simply the sum of the individual emf's of the individual cells. -- Even if one cell is connected backwards in the string. Then its emf is considered negative when the sum is being performed. -- All of this is true only as long as there is no external connection between the ends of the stack, you're measuring the emf on an 'open-circuit' basis with a voltmeter, and the cells are not providing any current to an external circuit. Once the series combination of cells is connected to an external circuit and begins to produce current, the total emf at the terminals of the stack will decrease. It'll depend on the magnitude of the current, and on the 'internal impedance' of each cell. If the cells are not precisely identical and in identical states of charge, then a calculation of the total emf is virtually impossible.
Why emf does not depend on resistance?
Because emf is the very source of voltage, either chemical or inductive, an can be meassured at open circuit only so, internal resistance of the supplier is not affecting it.
Why emf of battery driving potentiometer is greater than emf of the cell to be measured?
Bcoz the emf which is to be measured is less than emf of driving cell....
Why you perform open circuit characteristics of a dc shunt generator?
to find the relation between emf generated and Field current .....
How do you determine the emf and internal resistance of a cell?
You can measure the emf of a cell by using a voltmeter, as this draws current from a cell. You can use the voltage, the emf, and the load resistance to determine the internal resistance of the cell.
Does adding electrons to the circuit increase the emf?
No. But if you increased the EMF across the circuit, then more electrons would flow through it each second.
How to measure the electromotive force (EMF) of a cell accurately?
To measure the electromotive force (EMF) of a cell accurately, use a voltmeter with high precision and connect it to the terminals of the cell. Ensure the cell is not connected to any external circuit during the measurement to prevent errors. Take multiple readings and calculate the average to minimize inaccuracies.
Why cant you use voltmeter 2 calculate emf of a cell?
A voltmeter measures potential difference across a component, which may not necessarily be equal to the EMF of a cell due to internal resistance in the cell and voltage drops across other components in the circuit. To accurately measure the EMF of a cell, a potentiometer or a high-resistance voltmeter is used in conjunction with a null point method.
Does the induced emf in a circuit depend on the resistence of the circuit?
The induced emf in a circuit is not directly dependent on the resistance of the circuit. It is primarily determined by the rate of change of magnetic flux through the circuit. However, the resistance of the circuit can affect the current flow and ultimately impact the magnitude of the induced emf through Ohm's law (V = IR).
Can the potential difference across a battery be greater than its EMF?
If the emf of a battery is E Volt, the potential difference across a battery is given byV = E -I r where I is the current in the circuit and r is the inetrnal resistance.Hence E and V will be equal only when I = 0.The maximum potential difference across the battery will be equal to E only if I = 0.In gnereral potential difference can be equal or less than the emf.E.m.f can never exceed the potential difference.=====================================A battery charger is a device used to put energy into a secondary cell or (rechargeable) batteryby forcing an electric current through it.Hence to charge a battery another source of emf is needed.The combined emf is now will be (E - E1) where E is the emf of the battey in quesiton and E 1 is the emf of the external source used to charge the battery.Note that E-E1 will be negative in sign.======================================...A battery is charged only when its emf is less than its maximum emf.Suppose that the maximum emf of a cell is 1.5V. The battery should be charged only when its emf is less than 1.5 V say 0.5 V.To charge the cell we use a different source of emf E1 say 3V.The positive of the second source is connected to the negative of the cell so that theCombined emf is now 0.5 - 3 = -2.5V.The negative sign indicates that the emf is opposite to the emf of the cell which is 0.5V.Since the cell is getting charged, the difference in emf is gradually reduced to zero when the cell is fully charged.In modern charging units there are provisions so that the cell is never allowed to be over charged, even if the charging unit is in on for about 12 hours.When the cell is fully charged, (that is when the emf of the cell is now 1.5V), the potential difference between either the second source or cell will be zero.Taking into consideration the sign of the emf and the direction of current through the cell and the sign of the potential difference, the potential difference will be always less than the emf of the cell (which gradually increases while charging).Note that the potential difference is negative if the emf of the cell is taken as positive.Also note that the cell is charged only when its emf is less than its maximum e.m.f
