$\"\"$

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Step 1:

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A DC generator has emf of 75 V.

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Internal resistance of the DC generator is 0.5 $\"\"$ .

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A battery has emf of 41 V.

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Internal resistance of the battery is 0.3 $\"\"$ .

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DC generator and battery are connected in parallel to the load of 2.5 $\"\"$ .

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Draw a circuit with above specifications.

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$\"\"$

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Find the current through generator :

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Redraw the circuit with current directions and nodes.

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$\"\"$

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Apply Junction rule at node b.

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$\"\"$ $\"\"$ .

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Step 2:

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Apply KVL to a loop abeda.

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$\"\"$

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Apply KVL to a loop bcfeb.

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$\"\"$

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Solve the equations (2) and (3).

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Multiply equation (2) with 0.3.

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$\"\"$

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Multiply equation (3) with 0.8.

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$\"\"$

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Subtract the equations (4) and (5).

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$\"\"$

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$\"\"$

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$\"\"$ A.

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Substitute $\"\"$ in equation (2).

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$\"\"$

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$\"\"$

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So the value of current through the generator is 50 A.

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The direction of the current is toward the node b.

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Step 3:

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(b)

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The value of current through the battery is $\"\"$ .

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$\"\"$

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So the value of current through the battery is 30 A.

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The direction of the current is toward the node c. (In opposite direction with respect to generator)

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Step 4:

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Find the potential difference across the load.

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The current across the load is $\"\"$ A.

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Use ohms law : $\"\"$ .

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Where i is current through load,

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R is resistance offered by load.

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$\"\"$

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The voltage across the load is 50 V.

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Solution:

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(a)

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The value of current through the generator is 50 A.

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The direction of the current is toward the node b.

\

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(b)

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The value of current through the battery is 30 A.

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The direction of the current is toward the node c

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(c)

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The voltage across the load is 50 V.

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\

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$\"\"$

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(2.2)

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Step 1:

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A 3 $\"\"$ resistor is connected in parallel with resistor of 3 $\"\"$ .

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This is connected in series to resistor R.

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The applied voltage to the network is 45 V.

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The current through the circuit is 10 A.

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Draw a circuit with above specifications.

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$\"\"$

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R₁ parallel to R₂ then the parallel resistance is

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$\"\"$

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$\"\"$ is in series with the R, hence equivalent resistance is

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$\"\"$

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Find the equivalent resistance.

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Equivalent resistance : $\"\"$ .

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$\"\"$

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$\"\"$ .

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

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$\"\"$

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The series resistance is 3 $\"\"$ .

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Step 2:

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(2.2.1)

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Find the voltage across the resistors.

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Current in a series network is same.

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Equilvalent parallel resistance is $\"\"$ .

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Voltage drop across the parallel as $\"\"$ .

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$\"\"$

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Voltage drop across the parallel network is $\"\"$ .

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Voltage drop across a resistor in parallel network is same.

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Hence Voltage drop each 3 $\"\"$ resistor is 15 V.

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Step 3:

\

Voltage drop across series resistance R.

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Voltage drop across the series resistance is $\"\"$ .

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$\"\"$

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Voltage drop across the series resistance is $\"\"$ .

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Solution :

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Voltage drop each 3 $\"\"$ resistor in parallel network is 15 V.

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Voltage drop across the series resistance is 30 V.

\

\

\

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(2.2.2)

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Step 1:

\

Find the current through the parallel resistors.

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Voltage drop across a resistor in parallel network is same.

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Voltage drop across each 3 $\"\"$ resistor in parallel network is 15 V. (From 2.2.1)

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

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$\"\"$

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Current through the resistance $\"\"$ is 5 A.

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Similarly current through the resistance $\"\"$ is 5 A.

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Solution ;

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Current through the resistance $\"\"$ and $\"\"$ is 5 A.