JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 23)
In a series LCR resonant circuit, the quality factor is measured as 100. If the inductance is increased by two fold and resistance is decreased by two fold, then the quality factor after this change will be __________.
Respondre
283
Explicació
Quality factor = $${{{X_L}} \over R} = {{\omega L} \over R}$$
$$Q = {1 \over {\sqrt {LC} }}{L \over R}$$
$$Q = \left( {{1 \over {\sqrt C }}} \right){{\sqrt L } \over R}$$
$$Q = {{XL} \over R} = {{\omega L} \over R} = {1 \over {\sqrt {LC} }}{L \over R} = {1 \over R}{{\sqrt L } \over {\sqrt C }}$$
$$Q' = {{\sqrt {2L} } \over {\left( {{R \over 2}} \right)\sqrt C }} = 2\sqrt 2 Q$$
Q' = 282.84
$$Q = {1 \over {\sqrt {LC} }}{L \over R}$$
$$Q = \left( {{1 \over {\sqrt C }}} \right){{\sqrt L } \over R}$$
$$Q = {{XL} \over R} = {{\omega L} \over R} = {1 \over {\sqrt {LC} }}{L \over R} = {1 \over R}{{\sqrt L } \over {\sqrt C }}$$
$$Q' = {{\sqrt {2L} } \over {\left( {{R \over 2}} \right)\sqrt C }} = 2\sqrt 2 Q$$
Q' = 282.84