Question

An inductor of 0.5 H and $200\Omega$ resistor are connected in series with A.C. source of 230 V and frequency 50 Hz, then calculate (1) Maximum current in the inductor (2) Phase difference between current and voltage and time difference (time lag)

• A. $I_{max}=1.28A,{38}^0,8^{\prime }and2.1ms$
• B. $I_{max}=1A,{36}^0,8^{\prime }and2.1ms$
• C. $I_{max}=1.28A,{38}^0,8^{\prime }and24.2ms$
• D. $I_{max}=1A,{38}^0,8^{\prime }and2.1ms$

Answer 1

Answer: (A)

$I_{max}=1.28A,{38}^0,8^{\prime }and2.1ms$

Given,

$L=0.5$, Supply$=230v,50H_2$

$R=200\omega$

Inductive reaction

$X_L=\omega L$

$\omega =2\pi f$

$\omega =2\pi \times 50=100\pi$

$X_L=100\pi \times 0.5=50\pi$

Total impendence$z=\sqrt{R^2\times x_2^2}$

$=\sqrt{(200{)}^2+(50\pi {)}^2}$

$z=254.31\omega$

(I) $i_{max}=?$

Given $v_{rms}=230$

$\Rightarrow v_{max}=\sqrt{2}{v}_{rms}=\sqrt{2}\times 230$

$i_{max}=\frac{v_{max}}{2}=\frac{230\sqrt{2}}{254.31}=1.28A$

(II) Phase difference

$\Rightarrow \cos \varphi =\frac{R}{z}$ (power factor)

$=\varphi ={\cos }^{-1}\frac{R}{z}={\cos }^{-1}\left(\frac{200}{254.31}\right)\approx {38}^o$

(III) Time lag

$\Delta t=\frac{\varphi }{360\times f}=\frac{{38}^o}{360\times 50}=2.1ms$.