Question

In the circuit shown below:

$(v_1+v_2)=\left[1sin\right(2\pi 10000t)+1sin(2\pi 30000t\left)\right]V$

The RMS value of the current through the resistor R will be minimum if the value of the capacitor C in microfarad is

• A. 0.28

Answer 1

The correct option is A 0.28

$V_1=1sin\left(2\pi 10000t\right)V$

$V_2=1sin\left(2\pi 30000t\right)V$

$\therefore f_1=10000Hz$

$f_2=30000Hz=3f_1$

For LC - 2

${\omega }_0=\frac{1}{\sqrt{LC}}\Rightarrow f_0=\frac{1}{2\pi \sqrt{LC}}$

$f_0=\frac{1}{2\pi \sqrt{100\times {10}^{-6}\times 2.53\times {10}^{-6}}}$

$So,for{f}_1=f_0{\mid }_{LC-2}\Rightarrow Resonance$

It act as open so I = 0

for LC - 1 to be open circuit i.e, for resonance

$f_2=3f_1=\frac{1}{2\pi \sqrt{LC}}$

$C=\frac{1}{(2\pi {)}^2\times (3f_1{)}^2\times L}$

$=\frac{1}{(2\pi {)}^2\times (30,000{)}^2\times 100\times {10}^{-6}}$

$=0.28\mu F$