A) \[i(t)={{i}_{0}}+{{i}_{1}}{{e}^{-t/\tau }}\]; \[\tau =3C\times \frac{R}{3}\]
B) \[i(t)={{i}_{0}}+{{i}_{1}}{{e}^{-t/\tau }}+{{i}_{2}}{{e}^{-t/2\tau }}\]; \[\tau =RC\]
C) \[i(t)={{i}_{1}}+{{i}_{1}}{{e}^{-t/\tau }}\]; \[\tau =3C\times \frac{R}{3}\]
D) \[i(t)={{i}_{0}}+{{i}_{1}}{{e}^{-t/\tau }}\]; \[\tau =3RC\]
Correct Answer: B
Solution :
[b] The three branches of the circuits carry currents \[i={{i}_{0}}\], \[i={{i}_{1}}\,{{e}^{t/RC}}\] and \[i={{i}_{2}}\,2{{e}^{-t/2Rc}}\] respectively. The current through the cell, i(t) can be found by using Kirchhoff's current law (or mode law).
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