Question

# A resistance of $$10\Omega$$ and a inductance of $$100mH$$ are connected in series through an alternating voltage source $$V=100\cos { \omega t }$$. Fond out phase difference between current and voltage for the circuit.

Solution

## Given: $$R=10\Omega$$ and Inductance (L)$$=100mH$$Voltage source $$==100\cos { \left( 100t \right) }$$$$V={ V }_{ 0 }\cos { \omega t } ;\{ { V }_{ 0 }=100V;\omega =100rad/s\quad$$$$\quad \therefore \tan { \phi } =\cfrac { { X }_{ L } }{ R } =\cfrac { \omega L }{ R } =\cfrac { 100\times 100\times { 10 }^{ -3 } }{ 10 }$$$$\tan { \phi } =1\Rightarrow \phi =\cfrac { \pi }{ 4 } ={ 45 }^{ o }$$PhysicsNCERTStandard XII

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