An inductor and a resistor are connected in s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For an $L-R$ circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + (L\omega)^2}$. The current is $I = V/Z$, so $R^2 + L^2\omega^2 = (V/I)^2$.
C

Vedclass · Accepted Answer

$60 \,mH, \sqrt{71} \,\Omega$

Vedclass · Answer

(D) For an $L-R$ circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + (L\omega)^2}$. The current is $I = V/Z$, so $R^2 + L^2\omega^2 = (V/I)^2$.
Case $1$: $I_1 = 0.5 \,A$, $f_1 = 175/\pi \,Hz$, $\omega_1 = 2\pi f_1 = 350 \,rad/s$, $V = 8\sqrt{2} \,V$.
$R^2 + L^2(350)^2 = (8\sqrt{2} / 0.5)^2 = (16\sqrt{2})^2 = 256 	imes 2 = 512$.
Case $2$: $I_2 = 0.4 \,A$, $f_2 = 225/\pi \,Hz$, $\omega_2 = 2\pi f_2 = 450 \,rad/s$, $V = 8\sqrt{2} \,V$.
$R^2 + L^2(450)^2 = (8\sqrt{2} / 0.4)^2 = (20\sqrt{2})^2 = 400 	imes 2 = 800$.
Subtracting the two equations: $L^2(450^2 - 350^2) = 800 - 512 = 288$.
$L^2(202500 - 122500) = 288 \Rightarrow L^2(80000) = 288$.
$L^2 = 288 / 80000 = 0.0036 \Rightarrow L = 0.06 \,H = 60 \,mH$.
Substituting $L$ into the first equation: $R^2 + (0.06 	imes 350)^2 = 512$.
$R^2 + (21)^2 = 512 \Rightarrow R^2 + 441 = 512$.
$R^2 = 71 \Rightarrow R = \sqrt{71} \,\Omega$.

Similar Questions

If $L$ and $R$ are respectively the inductance and resistance,then the dimensions of $\frac{R}{L}$ will be

In an $ac$ circuit,the potential difference across an inductance and resistance joined in series are respectively $16\, V$ and $20\, V$. The total potential difference across the circuit is.......$V$

When $100 \, V$ $DC$ is supplied across a solenoid,a current of $1.0 \, A$ flows in it. When $100 \, V$ $AC$ is applied across the same coil,the current drops to $0.5 \, A$. If the frequency of the $AC$ source is $50 \, Hz$,then the impedance and inductance of the solenoid are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module