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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 $$==100\cos { \left( 100t \right)  } $$
$$V={ V }_{ 0 }\cos { \omega t } ;\{ { V }_{ 0 }=100V;\omega =100rad/s\quad $$
$$\quad \therefore \tan { \phi  } =\cfrac { { X }_{ L } }{ R } =\cfrac { \omega L }{ R } =\cfrac { 100\times 100\times { 10 }^{ -3 } }{ 10 } $$
$$\tan { \phi  } =1\Rightarrow \phi =\cfrac { \pi  }{ 4 } ={ 45 }^{ o }$$
1763138_1836819_ans_9e096aa3a10d4572892aa64a7e3c05cd.png

Physics
NCERT
Standard XII

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