\"\"

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Step 1:

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Use Superposition principle to calculate volatge across A and B.

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Super Position Theorem :

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Take one current source at a time and replace all other with voltage with short or internal resistance and current sources with open.

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\"image\"

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The current flowing in the circuit is \"\".

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\"\" in polar form can be written as \"\".

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\"\"

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\"\" in complex form can be written as \"\".

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Apply current divider rule: \"\".

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Find the total impdenace \"\".

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\"\" is in series with \"\".

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\"\"

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\"\"

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\"\" is in series with \"\".

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\"\"

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\"\"

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\"\" is in parallel to \"\"

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\"\"

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\"\" is in series with \"\", \"\" and \"\"

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\"\"

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\"\"

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Current divider rule:

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\"\"

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\"\"

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Find the voltage drop across the impedance \"\".

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\"\"

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\"\"

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\"\" V.

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Step 2:

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Take one voltage source at a time and eplace all other with voltage with short or internal resistance and current sources with open.

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\"image\"

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The voltage in the circuit is \"\"

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\"\" in polar form can be written as \"\".

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\"\"

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\"\" in complex form can be written as \"\".

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Apply KVL to loop 1:

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\"\"

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Find the voltage drop across the impedance \"\".

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\"\"

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\"\"

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\"\" V.

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Step 3:

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By superposition principle : 

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The Voltage acoss the resistor \"\" is \"\".

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\"\"

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\"\"

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\"\"

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\"\"

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Voltage acoss the resistor \"\" is \"\"

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Solution:

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Voltage acoss the resistor \"\" is \"\"