To an ac power supply of 220 V at 50 Hz, a resistor…

Correct Answer: 7.8 A and $45^{\circ}$

Step I: Identify given values: $V_{rms} = 220~V$, $R = 20~\Omega$, $X_C = 25~\Omega$, $X_L = 45~\Omega$.
Step II: Calculate the net impedance $Z$: $Z = \sqrt{20^2 + (45 - 25)^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2}~\Omega \approx 28.28~\Omega$.
Step III: Calculate the current $I$: $I = \frac{V_{rms}}{Z} = \frac{220}{20\sqrt{2}} = \frac{11}{\sqrt{2}} \approx \frac{11}{1.414} \approx 7.78~A \approx 7.8~A$.
Step IV: Calculate the phase angle $\phi$: $\tan \phi = \frac{X_L - X_C}{R} = \frac{45 - 25}{20} = \frac{20}{20} = 1$.
Step V: Since $\tan \phi = 1$, the phase angle is $\phi = 45^{\circ}$.

सही उत्तर: 7.8 A और $45^{\circ}$

Step I: दिए गए मानों को पहचानें: $V_{rms} = 220~V$, $R = 20~\Omega$, $X_C = 25~\Omega$, $X_L = 45~\Omega$।
Step II: शुद्ध प्रतिबाधा $Z$ की गणना करें: $Z = \sqrt{20^2 + (45 - 25)^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2}~\Omega \approx 28.28~\Omega$।
Step III: धारा $I$ की गणना करें: $I = \frac{V_{rms}}{Z} = \frac{220}{20\sqrt{2}} = \frac{11}{\sqrt{2}} \approx \frac{11}{1.414} \approx 7.78~A \approx 7.8~A$।
Step IV: कला कोण $\phi$ की गणना करें: $\tan \phi = \frac{X_L - X_C}{R} = \frac{45 - 25}{20} = \frac{20}{20} = 1$।
Step V: चूँकि $\tan \phi = 1$ है, इसलिए कला कोण $\phi = 45^{\circ}$ है।