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

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 circuitPhase 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$$\displaystyle ={ I }_{ 0 }\sin { \phi } =\frac { { I }_{ 0 } }{ \sqrt { 2 } } =\frac { 1 }{ \sqrt { 2 } } .\frac { { E }_{ 0 } }{ Z }$$$$\displaystyle =\frac { 1 }{ \sqrt { 2 } } \frac { 220 }{ \sqrt { { X }_{ L }^{ 2 }+{ R }^{ 2 } } } =\frac { 1 }{ \sqrt { 2 } } .\frac { 220 }{ \sqrt { { 220 }^{ 2 }+{ 220 }^{ 2 } } }$$$$\displaystyle =\frac { 1 }{ 2 } =0.5A$$PhysicsNCERTStandard XII

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