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A cell of emf 2 V and internal resistanc...

A cell of emf 2 V and internal resistance `1Omega` has its terminals joined by resistanecs of 5 and `10Omega` in paralle. Using kirchhoff's laws, find the current through the cell and the current drawn through each resistance.

Text Solution

Verified by Experts

The correct Answer is:
`(6)/(13)A,(4)/(13)A,(2)/(13)A`
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A cell of emf 2V and internal resistance 4 Omega is connected across a parallel combination of two resistors of resistance 10 Omega and 20 Omega . Find the current through each resistor using Kirchhoff's laws.

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Knowledge Check

  • A cell of 2.1 V gives 0.2 A current through resistance 10Omega The internal resistance is:

    A
    `5 Omega`
    B
    `50 Omega`
    C
    `0.05Omega`
    D
    `0.5 Omega`

  • A cell of emf 2 V and internal resistance 0.1 Omega is connected with a resistance of 3.9 Omega The voltage across the cell terminal will be

    A
    `1.95V`
    B
    `1.90V`
    C
    `0.50V`
    D
    `2.00V`

  • A cell of emf 2.0 V and internal resistance 0.1 Omega is connected with a resistance of 3.9 Omega . The voltage across the cell terminals will be

    A
    0.50 V
    B
    1.90 V
    C
    1.95 V
    D
    2.0 V

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    Thirty six cells each of emf 1.5 V and internal resistance 0.5 Omega are used to send current through an external resistor of resistance 2 Omega . What is the best mode of grouping them and the current through the external resistor

    A cell of emf 1.1 V and internal resistance 0.5 Omega is connected to a wire of resistance 0.5 Omega . Another cell of the same emf is connected in series bur the current in the wire remain the same .Find the internal resistance of second cell

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