**Why is the terminal voltage less than EMF during discharging of battery?**

**Why is the terminal voltage less than EMF during discharging of battery?**

The equivalent circuit of the battery is as given below. A voltage source can be represented as EMF source with an internal series resistance.

**terminal voltage.**

_{L}R

_{s }is equal to zero because I

_{L }is zero, and in this condition the battery is said to be open circuited. In open circuited condition, the battery terminal voltage is equal to the EMF of the battery.

The cell voltage having EMF V is connected to the load R_{L}. The cell supplies current I_{L} to the load. According to Kirchhoff’s voltage law sum of the potential drops in a closed loop electric circuit is zero.

V = I_{L}R_{s} + I_{L}R_{L}

V = I_{L}R_{s} + (V_{T}/R_{L}) R_{L}

V = I_{L}R_{s} + V_{T }

_{If battery is open circuit,}I_{L = 0}

V =I_{L}R_{s} + I_{L}R_{L}

V=0 x R_{s} + (V_{T}/R_{L}) R_{L}

V=0 + V_{T }

_{VT =V}

When the battery is not connected to the load the load current is zero and battery EMF is equal to the terminal voltage.

Now if the load is connected to the battery, the battery starts supplying the current to the load.

V = I_{L}R_{s} + I_{L}R_{L}

V = I_{L}R_{s} + (V_{T}/R_{L}) R_{L}

V = I_{L}R_{s} + V_{T }

**V _{T }= V- I_{L}R_{s}**

From above, it is clear that the terminal V_{T} voltage is less than the EMF of battery. The reason of less terminal voltage when battery is under load is voltage drop inside the battery caused by battery internal resistance R_{s}.

The terminal voltage is further depends on the magnitude of the load. The load current depends on the value of the load resistance. The lower value of load resistance causes to draw more current from the battery, and consequently the more voltage drop takes place inside the battery, and there will be less terminal voltage. Let us understand this with illustrative example.

**Case 1: When battery is under no load**

When there is no current flowing through the battery, the terminal voltage is equal to the battery EMF. The battery EMF is 9 volts and the terminal voltage is equal to battery EMF that is 9 volts.

**Case 2: When battery is under load – Load is 9** **Ω**

When the battery of 9 volts EMF is connected to 9 Ω resistance the battery will supply 1.49 amp. current to the circuit and the terminal voltage is 8.94 volts. The voltage drop across the internal resistance of the battery is 0.015 volts.

**Case 3: When battery is under load – Load is 6 Ω**

When the battery of 9 volts EMF is connected to 6 Ω resistance the battery will supply 0.99 amp. current to the circuit and the terminal voltage is 8.91 volts.The voltage drop across the internal resistance of the battery is 0.001 volts.

From above, it is clear that voltage drop across the internal resistance increases with increase in the current.

When there is no current flowing through the battery, the terminal voltage is equal to the battery EMF. The battery EMF is 9 volts and the terminal voltage is equal to battery EMF that is 9 volts.

The terminal voltage decrease with increase of load current or the current through the battery.The graph showing relationship between voltage and current is as shown below.

**Illustrative Example:**

A battery has 9 volts emf and internal resistance of 0.1 Ohms. Calculate

a) Calculate battery terminal voltage when it is connected to 15 Ohms load.

b) The terminal voltage when it is connected to 1.0 Ohms load.

b) When battery is connected to 1.0 Ohms load