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Step 1:

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2.1.1)

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Two capacitors are in series.

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$\"\"$

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Substitute $\"\"$ microfarads, $\"\"$ microfarads.

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$\"\"$

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Total capacitance is 6 microfarads.

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Solution :

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Total capacitance is 6 microfarads.

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Step 1:

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2.1.2)

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From (2.1.1), total capacitance is 6 microfarads.

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Find the potential difference across each capacitance.

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Now find the total charge $\"\"$ .

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$\"\"$ .

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Substitute $\"\"$ and $\"\"$ in the total charge.

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$\"\"$

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The total charge is

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$\"\"$

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Find the voltage across the capacitance $\"\"$ .

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$\"\"$

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Find the voltage across the capacitance $\"\"$ .

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$\"\"$

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The voltage across the capacitors $\"\"$ are $\"\"$ and $\"\"$ .

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Solution :

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The voltage across the capacitor $\"\"$ is $\"\"$ .

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The voltage across the capacitor $\"\"$ is $\"\"$ .

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2.2

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Lenzs law states that when an emf is generated by a change in magnetic flux according to Faradayss Law , the polarity of the induced emf is such, that it produces an current whose magnetic field opposes the change which produces it.

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The negative sign used in Faradays law of electromagnetic induction, indicates that the induced emf $\"\"$ and the change in magnetic flux $\"\"$ have opposite signs.

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$\"\"$

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Where $\"\"$ is Induced emf,

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$\"\"$ is change in magnetic flux,

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$\"\"$ is no of turns in coil

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Step 1:

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Length of the aluminium is 22 m.

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The diameter of the aluminum wire is d₁= 4 mm.

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Since the length are same, the length of the copper wire is 22 m.

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Specific resistivity of the aluminum is $\"\"$ .

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Specific resistivity of the copper $\"\"$ .

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Total current in the circuit is i_t = 25 A.

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Current through copper wire is i₂ = 15 A.

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The aluminum wire and copper wire are connected in parallel.

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Total Current = i₁ + i₂ .

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25 = i₁ + 15

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i₁ =25 - 15

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i₁ = 10 A.

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Current through aluminum wire is i₁ = 10 A.

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Step 2:

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Law of Resistivity:

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Resistance offered by a conductor is given by $\"\"$ .

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Resistance offered by aluminum wire is $\"\"$ .

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Area of the aluminum wire is $\"\"$ .

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Resistance offered by aluminum wire is $\"\"$ .

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Resistance offered by copper wire is $\"\"$ .

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Area of the copper wire is $\"\"$ .

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Resistance offered by copper wire is $\"\"$ .

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Ratio of the resistance is

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$\"\"$

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Step 3:

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In a parallel combination, the voltage across the element are same.

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$\"\"$

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$\"\"$

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Substitute $\"\"$ in equation (1).

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$\"\"$

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Substitute the corresponding values in the above formula.

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$\"\"$

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The diameter of the copper wire is 3.817 mm.

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Step 1:

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(b)

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Specific resistivity of the aluminum $\"\"$ .

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Specific resistivity of the copper $\"\"$ .

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Voltage drop across the aluminum is $\"\"$ .

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Current through copper wire is i₂ = 15 A.

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Current through aluminum wire is i₁ = 10 A.

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The diameter of the aluminum wire is d₁ = 4 mm.

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The diameter of the copper wire is d₂ = 3.817 mm.

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Resistance offered by aluminum wire is $\"\"$ .

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$\"\"$

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Resistance offered by aluminum is $\"\"$ .

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Voltage drop across the aluminum:

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$\"\"$

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Voltage drop across the aluminum is 0.2229 v.

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In a parallel combination, the voltage across both the elements are same.

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Voltage drop across the copper is 0.2229 v.

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Solution:

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Voltage drop across the conductors is 0.2229 v.

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Step 1:

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A coil has a resistance 125 ohms at 50 C.

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Resistance of the coil increases 250 ohms at 300 C.

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Find the temperature coefficient of the conductor.

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$\"\"$

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Therefore the temperature coefficient is 0.004 per

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