JEE Advance - Physics (2011 - Paper 2 Offline - No. 15)
A series RC combination is connected to an AC voltage of angular frequency $$\omega$$ = 500 rad/s. If the impedance of the RC circuit is R$$\sqrt{1.25}$$, the time constant (in millisecond) of the circuit is __________.
Answer
4
Explanation
We have impedance in the circuit
$$Z = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}}} \right)}^2}} $$
However, $$Z = R\sqrt {1.25} $$ and $$R\sqrt {1.25} = \sqrt {{R^2} + {{\left( {{1 \over {\omega C}}} \right)}^2}} $$
$$\Rightarrow 0.25{R^2} = {1 \over {{{(\omega C)}^2}}}$$
The time constant is
$$RC = \sqrt {{1 \over {0.25 \times {{500}^2}}}} = 4$$ ms
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