An a.c. source of angular frequency is connected across a resistor R and a capacitor C in series

Question

An a.c. source of angular frequency $ω$ is connected across a resistor R and a capacitor C in series. The current is observed as $I$ . Now the frequency of the source is changed to $ω / 4$ , (keeping the voltage unchanged) the current is found to be $I / 3$ . The ratio of resistance to reactance at frequency $ω$ is [2026]

Smart Achievers · Accepted Answer

(3)

$As we know that current in RLC circuit is given by$

$ω / 4$

$I = \frac{V_{0}}{\sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}}$

$As circuit is RC$

$I_{0} = \frac{V_{0}}{{(R^{2} + X_{C}^{2})}^{1 / 2}}$

$\frac{I_{0}}{3} = \frac{V_{0}}{{[R^{2} + {(4 X_{C})}^{2}]}^{1 / 2}}$

$R^{2} + {(X_{C})}^{2} = \frac{R^{2} + 16 X_{C}^{2}}{9}$

$\frac{8 R^{2}}{9} = \frac{7 X_{C}^{2}}{9}$

$\frac{R}{X_{C}} = \sqrt{\frac{7}{8}}$

Solution

1 $\sqrt{\frac{6}{7}}$

2 $\sqrt{\frac{3}{5}}$

3 $\sqrt{\frac{7}{8}}$

4 $\sqrt{\frac{3}{4}}$

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Solution

1 67

2 35

3 78

4 34

Ans. (3) As we know that current in RLC circuit is given by I=VZ=V[R2+(XL-XC)2]1/2 I=V0R2+(ωL-1ωC)2 As circuit is RC I0=V0(R2+XC2)1/2 I03=V0[R2+(4XC)2]1/2 R2+(XC)2=R2+16XC29 8R29=7XC29 RXC=78

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1 $\sqrt{\frac{6}{7}}$

2 $\sqrt{\frac{3}{5}}$

3 $\sqrt{\frac{7}{8}}$

4 $\sqrt{\frac{3}{4}}$