The variation curve of charge on the plates of capacitor for \[{{t}_{0}}<t<2{{t}_{0}}\] w.r.t. time is
A) Straight line
B) Parabolic
C) Exponential curve
D) Logarithmic curve
Correct Answer: A
Solution :
[a] P.D. across resistance as a function of time \[{{v}_{R}}=\frac{{{v}_{0}}t}{{{t}_{0}}}(0\le t\le {{t}_{0}})\] \[{{v}_{R}}={{v}_{0}}\,\,({{t}_{0}}\le t\le 2{{t}_{0}})\] \[{{v}_{R}}=2{{v}_{0}}-\frac{{{v}_{0}}}{2{{t}_{0}}}t\,\,\,(2{{t}_{0}}\le t\le 6{{t}_{0}})\] \[{{v}_{R}}=-{{v}_{0}}\,\,\,(6{{t}_{0}}\le t\le 7{{t}_{0}})\] \[{{v}_{R}}=\frac{{{v}_{0}}}{{{t}_{0}}}t-8{{v}_{0}}\,\,\,\,\,\,\,(7{{t}_{0}}\le t\le 8{{t}_{0}})\] Current is a function of time is \[\frac{{{v}_{R}}}{R}\]. \[0\le t\le {{t}_{0}};\] \[i=\frac{{{v}_{0}}}{R{{t}_{0}}}t\] \[\frac{dq}{dt}=\frac{{{v}_{0}}}{R{{t}_{0}}}t\] \[\int\limits_{0}^{q}{dq=\frac{{{v}_{0}}}{R{{t}_{0}}}}\int\limits_{0}^{t}{t\,\,dt}\] \[q=\frac{{{v}_{0}}}{2R{{t}_{0}}}{{t}^{2}}\] (Parabolic curve) In the interval; \[{{t}_{0}}<t<2{{t}_{0}}\] \[I=\frac{{{v}_{0}}}{R}\] \[\int\limits_{\frac{{{v}_{0}}{{t}_{0}}}{2R}}^{q}{dq}=\frac{{{v}_{0}}}{R}\int\limits_{{{t}_{0}}}^{t}{dt}\] \[q=\frac{{{v}_{0}}{{t}_{0}}}{2R}+\frac{{{v}_{0}}}{R}(t-{{t}_{0}})\] (linear variation)
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