A) their brightness will be same
B) \[{{B}_{1}}\]will be brighter than\[{{B}_{2}}\]
C) as frequency of supply voltage is increased, brightness of\[{{B}_{1}}\]will increase and that of\[{{B}_{2}}\]will decrease
D) only\[{{B}_{2}}\]will glow because the capacitor has infinite impedance
Correct Answer: C
Solution :
\[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2\pi vC}\] \[=\frac{1}{2\pi \times 50\times 500\times {{10}^{-6}}}\] \[=\frac{1}{5\pi \times {{10}^{-2}}}=\frac{20}{\pi }=6.37\,\Omega \] \[{{X}_{L}}=\omega L=2\pi vL\] \[=2\pi \times 50\times 10\times {{10}^{-3}}\] \[=\pi \,\Omega =3.14\,\Omega \] As\[{{X}_{C}}>{{X}_{L}},\]so current in the lower branch will be higher, so\[{{B}_{2}}\]will glow brighter. On increasing the frequency\[{{X}_{C}}\]will decrease and \[{{X}_{L}}\]will increase, so brightness of\[{{B}_{1}}\]will increase and brightness of\[{{B}_{2}}\]will decrease.
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