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A battery of 6 V internal resistance 0.5...

A battery of 6 V internal resistance `0.5 Omega` is joined in parallel with another of 10 V and internal resistnace `1 Omega`. The combination sends a current through an external resistance of `12 Omega`. Find the current through each battery.

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To solve the problem, we need to analyze the circuit formed by the two batteries and the external resistor. We will use Kirchhoff's laws to find the currents through each battery. ### Step-by-Step Solution: 1. **Identify the Components:** - Battery 1 (E1): Voltage = 6 V, Internal Resistance (r1) = 0.5 Ω - Battery 2 (E2): Voltage = 10 V, Internal Resistance (r2) = 1 Ω - External Resistance (R) = 12 Ω ...
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A battery of emf 6V internal resistance 5 Omega is joined in parallel with another cell of emf of 10V and internal resistance 1Omega and the combination is used to send current through a resistor of 12Omega . Calculate the current through each battery.

A battery of emf 6 volts and internal resistance 1Omega is connected in parallel to another battery of emf 8V and internal resistance 2Omega . The combination is then used to send current through an external resistance of 10Omega . Find the current through the external resistance.

Knowledge Check

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

    A
    `0.5Omega`
    B
    `0.8 Omega`
    C
    `1.0 Omega`
    D
    `0.2 Omega`

  • The internal resistance of a 2.1 V cell which gives a current 0.2 A through a resistance of 10 Omega

    A
    `0.2 Omega`
    B
    `0.5 Omega`
    C
    `0.8 Omega`
    D
    `1.0 Omega`

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