NEET · Physics · STD 12 -7. Alternating current
To an ac power supply of 220 V at 50 Hz , a resistor of \(20 \Omega\), a capacitor of reactance \(25 \Omega\) and an inductor of reactance \(45 \Omega\) are connected is series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively-
- A \(7.8\text{ A}\) and \(30^{\circ}\)
- B \(7.8\text{ A}\) and \(45^{\circ}\)
- C \(15.6\) and \(30^{\circ}\)
- D \(15.6\) and \(45^{\circ}\)
Answer & Solution
Correct Answer
(B) \(7.8\text{ A}\) and \(45^{\circ}\)
Step-by-step Solution
Detailed explanation
\(X _{ C }=25 \Omega \)
\( X _{ L }=45 \Omega \)
\( Z =\sqrt{ R ^2+\left( X _{ L }- X _{ C }\right)^2} \)
\( Z =\sqrt{20^2+(45-25)^2} \)
\( Z =20 \sqrt{2} \)
\( I_{\text {rms }}=\frac{V_{\text {rms }}}{Z}=\frac{220}{20 \sqrt{2}}=\frac{11}{\sqrt{2}}=7.8 A \)
\( \tan \phi=\frac{X_L-X_C}{R}=\frac{45-25}{20}=\frac{20}{20}=1 \)
\( \phi=45^{\circ}\)
\( X _{ L }=45 \Omega \)
\( Z =\sqrt{ R ^2+\left( X _{ L }- X _{ C }\right)^2} \)
\( Z =\sqrt{20^2+(45-25)^2} \)
\( Z =20 \sqrt{2} \)
\( I_{\text {rms }}=\frac{V_{\text {rms }}}{Z}=\frac{220}{20 \sqrt{2}}=\frac{11}{\sqrt{2}}=7.8 A \)
\( \tan \phi=\frac{X_L-X_C}{R}=\frac{45-25}{20}=\frac{20}{20}=1 \)
\( \phi=45^{\circ}\)
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