Explain why the p.d. across the terminals of a cell is more in open circuit and reduced in a closed circuit??????
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An ideal battery or cell has no internal resistance, ie., r = 0.
So there is no voltage drop across the materials in side the battery, when it is connected in a circuit with some external elements.
When a battery is not connected in a circuit or it is open, then there is no current passing through it. So the emf is the potential difference observed by measurement.
When a battery has an internal resistance r, and it is connected through a circuit, there is a current I flowing in the circuit. Hence, there is a voltage drop across this internal resistance. So P.d. across battery = emf - r I.
So there is no voltage drop across the materials in side the battery, when it is connected in a circuit with some external elements.
When a battery is not connected in a circuit or it is open, then there is no current passing through it. So the emf is the potential difference observed by measurement.
When a battery has an internal resistance r, and it is connected through a circuit, there is a current I flowing in the circuit. Hence, there is a voltage drop across this internal resistance. So P.d. across battery = emf - r I.
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