Question

If emf $E=4\cos 1000t$ volt is applied to an L-R circuit of inductance $3$mH and resistance $4\Omega$, the amplitude of current in the circuit is.

• A. $\frac{4}{\sqrt{7}}A$
• B. $1.0A$
• C. $\frac{4}{7}A$
• D. $0.8A$

Answer 1

Answer: (D)

$0.8A$

As we know,

$Z=\sqrt{R^2+X_L^2}$
Here, $R=4\Omega ,X_L=L_{\omega }$
$=3\times {10}^{-3}\times 1000\Omega$
$=3\Omega$
Then, $Z=\sqrt{(4{)}^2+(3{)}^2}$
or $Z=\sqrt{16+9}=\sqrt{25}$
or $Z=5\Omega$
Hence, current, $I_0=\frac{E_0}{Z}$
$=\frac{4}{5}=0.8A$