Physics · Current Electricity · NEET
No. EMF (E) is the maximum voltage the cell provides in the open-circuit case, when no current is drawn (I = 0). Terminal voltage (V) is the voltage across the two terminals while a current I flows to an external resistor. They are equal only when I = 0. As soon as current flows, V = E - I r, so terminal voltage falls below EMF.
Every real cell has a small internal resistance r. When current I flows, a part of the EMF is used up inside the cell to push current through r. That lost part equals I r. So the voltage left for the outside circuit is V = E - I r, which is smaller than E. More current means a bigger I r drop and a lower terminal voltage.
Yes, but only when the cell is being charged. During charging, an outside source pushes current backward into the cell, so the internal I r drop adds instead of subtracts: V = E + I r. In normal discharge (the usual NEET case), V is always less than or equal to E.
A real voltmeter draws a small current, so it reads the terminal voltage V, not the true EMF. To read the true EMF you need a method that draws zero current from the cell, such as a potentiometer at its balance point. This is why a potentiometer measures EMF more accurately than a voltmeter.
Use V = E - I r for two different currents. If terminal voltage is V1 at current I1 and V2 at current I2, then E = (V1 I2 - V2 I1) / (I2 - I1) and r = (V1 - V2) / (I2 - I1). This two-equation trick appears often in NEET numericals.
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.
For a discharging cell, V = E - I r, where E is the EMF, I is the current, and r is the internal resistance. For a charging cell it becomes V = E + I r.
They are equal only in the open-circuit condition, when no current flows (I = 0). Then the I r drop is zero and V = E.
Both are measured in volts (V). EMF is energy given per unit charge by the source; terminal voltage is the potential difference across the cell's terminals in the circuit.
Yes. Since I = E/(R + r), a larger external resistance R gives a smaller current and a smaller I r drop, so the terminal voltage rises toward E. A very small R (near short circuit) gives large current and low terminal voltage.
A potentiometer at balance draws zero current from the cell, so there is no I r drop and it reads the true EMF. A voltmeter draws a small current, causing a tiny I r drop, so it reads terminal voltage instead.
