Question
Find the time constant (in $$\mu $$s) for the given $$RC$$ circuits in the given order respectively
Find the time constant (in $$\mu $$s) for the given $$RC$$ circuits in the given order respectively
$${R_1} = 1\Omega ,{R_2} = 2\Omega ,{C_1} = 4\mu F,{C_2} = 2\mu F$$
A.
$$18,4,\frac{8}{9}$$
B.
$$18,\frac{8}{9},4$$
C.
$$4,18,\frac{8}{9}$$
D.
$$4,\frac{8}{9},18$$
Answer :
$$18,\frac{8}{9},4$$
Solution :
KEY CONCEPT :
Time constant of $$R - C\,{\text{circuit}}\,{\text{is}}\,\tau = {R_{eq}}{C_{eq}}$$
$$\eqalign{ & \left( {\text{i}} \right)\,\,{\tau _1} = \left( {2 + 1} \right)\left( {2 + 4} \right) = 18\mu {\text{s}} \cr & \left( {{\text{ii}}} \right)\,{\tau _2} = \left( {\frac{{2 \times 1}}{{2 + 1}}} \right)\left( {\frac{{2 \times 4}}{{2 + 4}}} \right) = \frac{8}{9}\mu {\text{s}} \cr & \left( {{\text{iii}}} \right)\,{\tau _3} = \left( {\frac{{2 \times 1}}{{2 + 1}}} \right) \times \left( {4 + 2} \right) = 4\mu {\text{s}} \cr} $$
KEY CONCEPT :
Time constant of $$R - C\,{\text{circuit}}\,{\text{is}}\,\tau = {R_{eq}}{C_{eq}}$$
$$\eqalign{ & \left( {\text{i}} \right)\,\,{\tau _1} = \left( {2 + 1} \right)\left( {2 + 4} \right) = 18\mu {\text{s}} \cr & \left( {{\text{ii}}} \right)\,{\tau _2} = \left( {\frac{{2 \times 1}}{{2 + 1}}} \right)\left( {\frac{{2 \times 4}}{{2 + 4}}} \right) = \frac{8}{9}\mu {\text{s}} \cr & \left( {{\text{iii}}} \right)\,{\tau _3} = \left( {\frac{{2 \times 1}}{{2 + 1}}} \right) \times \left( {4 + 2} \right) = 4\mu {\text{s}} \cr} $$
