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Question

A coil has an inductance of 0.7 henry and is joined in series with a resistance of $$\displaystyle 220\Omega $$ . When the alternating emf of 220 V at 50 Hz is applied to it then the phase through which current lags behind the applied emf and the wattless component of current in the circuit will be respectively 


A
30o,1A
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B
45o,0.5A
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C
60o,1.5A
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D
None of these
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Solution

The correct option is B $$\displaystyle { 45 }^{ o },0.5A$$
$$\displaystyle L=0.7H,R=220\Omega ,{ \varepsilon  }_{ 0 }=220V,v=50Hz$$

This is an L - R circuit
Phase difference,

$$\displaystyle \tan { \phi  } =\frac { { X }_{ L } }{ R } =\frac { \omega L }{ R } =\frac { 2\pi vL }{ R } $$

$$\displaystyle \left[ { X }_{ L }=2\pi vL=2\times \frac { 22 }{ 7 } \times 50\times 0.7=220\Omega  \right] $$

$$\displaystyle =\frac { 220 }{ 220 } =1\quad or,\quad \phi ={ 45 }^{ o }$$

Wattless component of current

$$\displaystyle ={ I }_{ 0 }\sin { \phi  } =\frac { { I }_{ 0 } }{ \sqrt { 2 }  } =\frac { 1 }{ \sqrt { 2 }  } .\frac { { E }_{ 0 } }{ Z } $$

$$\displaystyle =\frac { 1 }{ \sqrt { 2 }  } \frac { 220 }{ \sqrt { { X }_{ L }^{ 2 }+{ R }^{ 2 } }  } =\frac { 1 }{ \sqrt { 2 }  } .\frac { 220 }{ \sqrt { { 220 }^{ 2 }+{ 220 }^{ 2 } }  } $$

$$\displaystyle =\frac { 1 }{ 2 } =0.5A$$

Physics
NCERT
Standard XII

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