the EMF of a cell is always greater than its terminal voltage. why?
Asked by babita bhatt | 18th Feb, 2014, 10:00: PM
EMF is the potential difference across the two terminals of a voltage source when it is not connected to any circuit. When the current starts to flow in the circuit, then there is a voltage drop due to internal resistance. Therefore, once it is connected to a circuit, the voltage reduces slightly because of the internal resistance of the voltage source.
The terminal voltage of the voltage source is the potential difference across the terminals when it is connected to a circuit. This is why the terminal voltage is always less than the EMF of the same voltage source.
Example:
Let,
EMF of the battery = 6 v
External resistance = 1.5 ohms
Internal resistance of cell = 0.5 ohms
Total resistance of the circuit r = 1.5 + 0.5 = 2.0 ohms
Current I = V/R = 6/2 = 3 A
Voltage drop across internal resistance V = IR = 3 x 0.5 = 1.5 V
Hence the voltage available to external circuit is = 6 – 1.5 = 4.5 V
EMF is the potential difference across the two terminals of a voltage source when it is not connected to any circuit. When the current starts to flow in the circuit, then there is a voltage drop due to internal resistance. Therefore, once it is connected to a circuit, the voltage reduces slightly because of the internal resistance of the voltage source.
The terminal voltage of the voltage source is the potential difference across the terminals when it is connected to a circuit. This is why the terminal voltage is always less than the EMF of the same voltage source.
Example:
Let,
EMF of the battery = 6 v
External resistance = 1.5 ohms
Internal resistance of cell = 0.5 ohms
Total resistance of the circuit r = 1.5 + 0.5 = 2.0 ohms
Current I = V/R = 6/2 = 3 A
Voltage drop across internal resistance V = IR = 3 x 0.5 = 1.5 V
Hence the voltage available to external circuit is = 6 – 1.5 = 4.5 V
Answered by | 19th Feb, 2014, 12:12: PM
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