Question

In the circuit diagram shown, $X_C=100\Omega ,X_L=200\Omega$ & $R=100\Omega .$ The effective current through the source is:

• A. 2 A
• B. $2\sqrt{2}A$
• C. 0.5 A
• D. $\sqrt{0.4}A$

Answer 1

Answer: (B)

$2\sqrt{2}A$

$I_R=\frac{V}{R}=\frac{200}{100}=2A$

$I^{\prime }=\frac{V}{X_L-X_C}=\frac{200}{100}=2A$

$I=\sqrt{I{{}_R}^2+I^{\prime 2}}=2\sqrt{2}Amp.$