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Karnataka Board PUCPUC Science 2nd PUC Class 12

A coil of 0.01 henry inductance and 1 ohm resistance is connected to 200 volt, 50 Hz ac supply. Find the impedance of the circuit and time lag between max. alternating voltage and current.

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Question

A coil of 0.01 henry inductance and 1 ohm resistance is connected to 200 volt, 50 Hz ac supply. Find the impedance of the circuit and time lag between max. alternating voltage and current.

Short/Brief Note
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Solution

R = 1 Ω, L = 0.01 H, V = 200 V, v = 50 Hz.


Impedance of the circuit

`Z = sqrt(R^2 + X_L^2) = sqrt(R^2 + (2pivL)^2)`

= `sqrt((1)^2 + (2 xx 3.14 xx 50 xx 0.1)^2)`

= `sqrt(1 + 9.86) = sqrt(10.86)`

= 3.3 Ω

∴ For phase angle, tan `phi = Z/R`

`tan phi = X_L/R = (2pivL)/R = (2 xx 3.14 xx 50 xx 0.01)/1`

`tan phi` = 3.14

`phi` = tan–1 3.14 = 72°

Phase difference

`phi = (72 xx pi)/180^circ` rad

`phi` = 1.20 radian

Time lag between alternating voltage and current

`phi` = ωt

`t = phi/ω = (72 xx pi)/(180 xx 2 xx pi xx 50) = 1/250` sec

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Chapter 7: Alternating Current - MCQ I [Page 44]

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NCERT Exemplar Physics Exemplar [English] Class 12
Chapter 7 Alternating Current
MCQ I | Q 7.23 | Page 44

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