Physics · Current Electricity · NEET
No, EMF is not a force. The name is historical and misleading. EMF is the work done per unit charge (energy per coulomb) by the cell to move charge from the negative to the positive terminal inside the cell. Its SI unit is the volt (joule per coulomb), not the newton. NCERT states clearly: the emf is not a force; it is the voltage difference between the two terminals of a source in open circuit. Do not write its unit as newton in NEET.
No. EMF (E) is the total voltage the cell can supply with no current drawn. Terminal voltage (V) is what you actually measure across the terminals when current I flows through the circuit. Because the cell has internal resistance r, some voltage is lost inside as Ir. So V = E - Ir when the cell supplies current. Terminal voltage is always less than EMF while the cell is discharging.
When no current flows (open circuit, I = 0). Then V = E - Ir = E - 0 = E. This is why a good voltmeter (very high resistance, draws almost no current) reads a value close to the EMF, and why a potentiometer at balance (zero current from the cell) measures the true EMF, not the terminal voltage.
Yes, but only when the cell is being charged (current is forced INTO the cell from outside). Then V = E + Ir, so V is greater than E. During normal discharge (cell supplying current), terminal voltage is always less than EMF. NEET questions almost always use the discharge case V = E - Ir.
EMF is the cause: the energy given per unit charge by the source across the whole cell. Potential difference is the effect: the energy used per unit charge across a part of the external circuit (a resistor). EMF exists even in an open circuit; PD across a resistor exists only when current flows through it. EMF drives the current; PD is the drop it produces across external components.
The terminal voltage of a battery of emf 10 V and internal resistance 1 ohm, connected to an external resistance of 4 ohm, is:
A resistor is connected to a battery of emf 12 V and internal resistance 2 ohm. If the current is 0.6 A, the terminal voltage of the battery is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The volt (V), which equals one joule per coulomb (J/C). EMF is energy per unit charge, not a force, so it is never measured in newtons.
For a cell supplying current, E = V + Ir, where V is terminal voltage, I is current and r is internal resistance. Rearranged, terminal voltage V = E - Ir. In an open circuit (I = 0), E = V.
EMF is the cell's total available voltage with no current drawn (open circuit). Voltage (terminal voltage) is what you measure when current actually flows, which is lower because of the internal resistance drop Ir.
When current I flows, a part of the EMF is used up pushing charge through the cell's own internal resistance r. This lost voltage equals Ir, so the terminal voltage V = E - Ir is less than the EMF E.
Use a potentiometer. At its balance point no current is drawn from the test cell (I = 0), so it measures the true EMF, not the reduced terminal voltage. A voltmeter draws a small current, so it reads slightly less than EMF.
