Question

The two elements in the circuit given below are,


v(t) = 50 cos(5000t + 30°) V
i(t) = 10 cos (5000t - 23.13°) A

• A. R = 4 $\Omega$ and C = 0.4 mF
• B. R = 4.5 $\Omega$ and L = 0.6 mH
• C. R = 3 $\Omega$ and L = 0.8 mH
• D. R = 3 $\Omega$ and C = 0.8 mF

Answer 1

The correct option is C R = 3 $\Omega$ and L = 0.8 mH

Current is lagging voltage by 53.13${}^{\circ }$
i.e., $\varphi =53.13{}^{\circ }$

$\therefore$ The elements will be R and L

$\tan \varphi =\frac{\omega L}{R}$

$\tan \left(53.13{}^{\circ }\right)=1.33=\frac{5000L}{R}...\left(i\right)$

[Since $\omega$ = 5000 rad/sec]

$\frac{V}{I}=|Z|=\frac{50}{10}=\sqrt{R^2+{\left(\omega L\right)}^2}$

$5=\sqrt{R^2+{\left(\omega L\right)}^2}$

$25=R^2+{\left(5000L\right)}^2$
From equation (i), ... (ii)

$L=\frac{R\times 1.33}{5000}$

$25=R^2+{\left(1.33R\right)}^2$

$R=3\Omega$
Substituting in equation (ii), we get

$L=\frac{3\times 1.33}{5000}=0.8mH$