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Step 1:

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The emf of the battery $\"image\"$ is 8V.

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The internal resistance of the battery is $\"\"$ .

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Two coils of their resistances $\"\"$ connected in parallel.

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Draw a circuit with above specifications :

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$\"\"$

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Step 2:

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Find the equivalent resistance :

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Resistor $\"\"$ parallel.

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Let $\"\"$ is equivalent parallel resistance.

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$\"\"$

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$\"\"$

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Now $\"\"$ is in series with the internal resistance of the battery.

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$\"\"$

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$\"\"$

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Step 3:

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Find the current in the circuit :

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From ohms law : $\"\"$ .

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$\"\"$

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Substitute $\"\"$ .

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$\"\"$

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Step 4:

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(1)

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Suppose that the battery with emf $\"image\"$ and internal resistance $\"image\"$ supplies a current $\"image\"$ through an external load resistor $\"image\"$ , then the potential difference across the battery is $\"\"$ .

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Substitute , $\"\"$ and $\"\"$ in $\"\"$ .

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$\"\"$

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The potential difference across battery terminal is $\"\"$ .

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Step 5:

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(2)

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The voltage in the parallel loop is $\"\"$ .

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The voltage in the parallel loop same.

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So, the voltage across each coil is same.

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Current through the coil which has resistance $\"\"$ .

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From ohms law : $\"\"$ .

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$\"\"$

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Current through the coil which has resistance $\"\"$ .

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From ohms law : $\"\"$ .

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$\"\"$

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Step 6:

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(3)

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Power observed by each coil :

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Power can be calculated using the relation : $\"\"$ .

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Where $\"\"$ is power absorbed,

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$\"\"$ is current through the coil,

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$\"\"$ resistance of the coil.

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Power absorbed by the $\"\"$ coil is

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$\"\"$

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Power absorbed by the $\"\"$ coil is

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$\"\"$

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Solution:

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The potential difference across battery terminal is $\"\"$ .

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Current through the $\"\"$ coil is $\"\"$ .

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Current through the $\"\"$ coil is $\"\"$ .

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Power absorbed by the $\"\"$ coil is 4.98 W.

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Power absorbed by the $\"\"$ coil is 3.32 W.

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