Question

A resistance and an inductance are connected in series across an $AC$ voltage, $V=100\sin \left(10t\right)$. The current in this circuit is given by, $I=4\sin (10t-\pi /4)$. The values of the inductance and resistance are, respectively,

• A. $L=1\text{H},R=10\Omega$
• B. $L=1.77\text{H},R=17.7\Omega$
• C. $L=0.1\text{H},R=1\Omega$
• D. $L=0.17\text{H},R=1.7\Omega$

Answer 1

The correct option is B $L=1.77\text{H},R=17.7\Omega$

Here, $V=100\sin \left(10t\right)\text{and}I=4\sin (10t-\pi /4)$

In $L-R$ series circuit, current lags the voltage by an angle $\varphi$, given by,

$\tan \varphi =\left(\frac{X_L}{R}\right)$

Here, $\varphi =\frac{\pi }{4}$

$\therefore X_L=R$ or $\omega L=R$

Here, $\omega =10\text{rad/s}$

$\therefore R=10L...\left(i\right)$

Further, $V_0=i_0|Z|$

$\therefore 100=4\sqrt{R^2+X_L^2}$

$\Rightarrow R^2+(\omega L{)}^2={\left(\frac{100}{4}\right)}^2$

$\Rightarrow 2R^2=625$ (as $\omega L=R$)

$\therefore R=17.7\Omega$

Using equation $\left(i\right)$,

$L=1.77\text{H}$

Hence, $\left(B\right)$ is the correct answer.