Courses
Courses for Kids
Free study material
Offline Centres
More
Store

A series RLC circuit has a bandwidth of 300rad/sec at a resonance frequency of 3000rad/sec when excited by a voltage source of 100V. The inductance of the coil is 0.1H. The value of R and voltage across C are respectively\begin{align} & \left( 1 \right)10\Omega and 100V \\ & \left( 2 \right)30\Omega and 100V \\ & \left( 3 \right)30\Omega and 1000V \\ & \left( 4 \right)300\Omega and 1000V \\ \end{align}

Last updated date: 19th Jun 2024
Total views: 393.9k
Views today: 4.93k
Verified
393.9k+ views
Hint: Here the bandwidth is the ratio of the given resistance and the inductance in the circuit. Hence by rearranging the equation we get the equation of R. Then by substituting the values of the bandwidth and the inductance we will the resistance. The quality factor is the ratio of resonant frequency to the band width. The product of the quality factor and source voltage is equal to the voltage across capacitance which is also equal to the voltage across inductance.

Given that V=100V
L=0.1H
In series RLC circuit,
$Bandwidth=\dfrac{R}{L}=300$
$\Rightarrow$ $R=Bandwidth\times L$
$\Rightarrow R=300\times 0.1$
$\therefore R=30\Omega$
$Q=\dfrac{\operatorname{Re}sonantFrequency}{bandwidth}$
where, Q is the quality factor.
\begin{align} & \Rightarrow Q=\dfrac{3000}{300} \\ & \therefore Q=10 \\ \end{align}
At resonance,
$\left| {{V}_{L}} \right|=\left| {{V}_{C}} \right|=Q{{V}_{S}}$
\begin{align} & \Rightarrow Q{{V}_{S}}=10\times 100 \\ & \therefore \left| {{V}_{L}} \right|=\left| {{V}_{C}} \right|=1000V \\ \end{align}

Hence option (3) is correct.