Question

A 100$\Omega$ resistance is connected in series with a 4H inductor. The voltage across the resistor is $V_R=2sin\left(1000t\right)V$. The voltage across the inductor is:

• A. $80sin(1000t+\frac{\pi }{2})$
• B. $40sin(1000t+\frac{\pi }{2})$
• C. $80sin(1000t-\frac{\pi }{2})$
• D. $40sin(1000t-\frac{\pi }{2})$

Answer 1

Answer: (A)

$80sin(1000t+\frac{\pi }{2})$

$V_R=2sin\left(1000t\right)V$ So, $V_{max}\left(R\right)=2V$

So, $w=1000$

$2\pi f=1000$

$f=\frac{500}{\pi }Hz$

$X_L=wL$

$=1000\times 4$

$=4000\Omega$

$R=100\Omega$

$i_R=\frac{V_R}{R}=\frac{2}{100}=2\times {10}^{-2}A=i_L$

($\because$current is same in series)

$V_{max}\left(L\right)=i_L\times X_L$

$=4000\times 2\times {10}^{-2}$

$=80V$

So, $V_L=80sin(1000t+\frac{\pi }{2})$

Since in inductor voltage lead by $\frac{\pi }{2}$