0
What else can I help you with?
What formula to find the emf and internal resistance for cell when given the current and the voltage?
I don't think you can do that, with the information provided.
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.
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.
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.
What is the difference between the emf of a battery and the lost volts when discharging called?
Internal resistance
What are the applications of comparing Emf of two cells with the help of potentiometer?
I think you are talking about it's internal resistance or it's ability to supply a current to a load without significant drop in voltage. Experiment to find EMF of unknown cell The 1 Metre scaled potentiometer is placed across a cell of known EMF and the 'jockey' or slider connected to one terminal of a galvanometer. The second cell is wired with similar polarity to the first cell and connected from the other side of the galvo to the common negative terminal of the two cells. The slider of the potentiometer is progressively moved up from zero until the galvo shows no current. The EMF of the unknown cell is now a direct proportion to the reading in cm read of the metre scale. Example If the known cell has EMF of 3 volts and the galvo balances at 50cm then the unknown cell has EMF of1.5Volts Dry cells used in flashlights have a fairly high internal resistance whereas a Lead acid car cell has a very low internal resistance allowing the starter motor to draw hundreds of Amperes without volt drop.
An old cell with an emf of 1.6 V has an internal resistance of 1.4 How much current will initially flow if its terminals are short-circuited?
I=e/r =1.6/1.4=1.14Amps
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.
Is potential difference always less than emf of the cell?
No. Because during charging process of a battery current flows in opposite direction to the discharging/consumption. so equation Emf=P.d. +Ir is changed to Emf=p.d. +Ir. Hence during charging process of a battery Potential difference is greater than electromotive force.
What is the internal resistance of a new D battery with an emf of 1.5 V and a wire of negligible resistance is connected between the terminals of the battery a current of 27 A is produced?
"In short, it is 0.055555555555555555555555555555556OhmsYou can use ohms law to calculate this.You will use the formula: Resistance Equals Voltage Divided by Current.It is written: R=V/ISo use the numbers you provided, and the formula above.1.5 / 27 = 0.055555555555555555555555555555556"I dispute this answer. emf = I(R+r) is the actual equation u need. From this equation, u will find that V = emf - Ir. Ohm's law is not applicable for this situation. However I do think you will need a voltage to find the internal resistance.
Why emf falls even when the cell is not in use?
When a cell is not in use, there can still be a small amount of current flowing through it due to internal factors like self-discharge or leakage currents. This internal current can lead to a drop in the cell's electromotive force (EMF) over time, even when the cell is not actively powering a device.
A cell of emf E volt and its internal resistance is 'r' ohm. The cell is connected to a load of 'R' ohm. The potential difference across the terminals of the cell is?
The resistance is equally proportionate to "r" in the case that it is above 1. Assuming "r" is greater than 1, the resistance is 4/3 multiplied by omega (the equal proportionate value for mass times ohms). If "r" is less than or equal to 1, there is no resistance.
