Question

a cell of emf 10volts and internal resistance 1ohm is connected across an external resistance of 4ohm the terminal potential of the cell will be:-

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Answer 1

Answer:

VERIFIED ANSWER :

Emf of the battery E=10 volts

Emf of the battery E=10 voltsInternal resistance of the cell r=3Ω

Emf of the battery E=10 voltsInternal resistance of the cell r=3ΩCurrent flowing through the circuit I=0.5A

Emf of the battery E=10 voltsInternal resistance of the cell r=3ΩCurrent flowing through the circuit I=0.5APotential drop across internal resistance V=Ir

Emf of the battery E=10 voltsInternal resistance of the cell r=3ΩCurrent flowing through the circuit I=0.5APotential drop across internal resistance V=Ir∴ V=0.5×3=1.5 volts

Emf of the battery E=10 voltsInternal resistance of the cell r=3ΩCurrent flowing through the circuit I=0.5APotential drop across internal resistance V=Ir∴ V=0.5×3=1.5 voltsThus terminal voltage of the battery V

Emf of the battery E=10 voltsInternal resistance of the cell r=3ΩCurrent flowing through the circuit I=0.5APotential drop across internal resistance V=Ir∴ V=0.5×3=1.5 voltsThus terminal voltage of the battery V 1

Emf of the battery E=10 voltsInternal resistance of the cell r=3ΩCurrent flowing through the circuit I=0.5APotential drop across internal resistance V=Ir∴ V=0.5×3=1.5 voltsThus terminal voltage of the battery V 1

Emf of the battery E=10 voltsInternal resistance of the cell r=3ΩCurrent flowing through the circuit I=0.5APotential drop across internal resistance V=Ir∴ V=0.5×3=1.5 voltsThus terminal voltage of the battery V 1 =E−V

Emf of the battery E=10 voltsInternal resistance of the cell r=3ΩCurrent flowing through the circuit I=0.5APotential drop across internal resistance V=Ir∴ V=0.5×3=1.5 voltsThus terminal voltage of the battery V 1 =E−V∴ V

Emf of the battery E=10 voltsInternal resistance of the cell r=3ΩCurrent flowing through the circuit I=0.5APotential drop across internal resistance V=Ir∴ V=0.5×3=1.5 voltsThus terminal voltage of the battery V 1 =E−V∴ V 1

Emf of the battery E=10 voltsInternal resistance of the cell r=3ΩCurrent flowing through the circuit I=0.5APotential drop across internal resistance V=Ir∴ V=0.5×3=1.5 voltsThus terminal voltage of the battery V 1 =E−V∴ V 1

Emf of the battery E=10 voltsInternal resistance of the cell r=3ΩCurrent flowing through the circuit I=0.5APotential drop across internal resistance V=Ir∴ V=0.5×3=1.5 voltsThus terminal voltage of the battery V 1 =E−V∴ V 1 =10−1.5=8.5 volts

Explanation:

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Answer 2

8.5 is the answer for question