A) 8V
B) 16V
C) 10 V
D) not possible to determine unless values of R and C are given
Correct Answer: B
Solution :
Key Idea: Voltage across capacitor lags behind current by \[I/{{c}^{2}}\]. The given circuit is a C-R series circuit, \[\sim \] is in phase with i, while \[\sim \] lags behind i by \[\sim \]. Hence, resultant potential is \[\sim \] Given, \[{{\left( \frac{2Gm}{r} \right)}^{1/2}}\le C\] \[{{\left( \frac{2gm}{r} \right)}^{1/2}}=C\] \[{{\left( \frac{2gm}{r} \right)}^{1/2}}\ge C\] \[{{\left( \frac{gm}{r} \right)}^{1/2}}\ge C\] \[\Delta V/V\] \[\frac{\Delta V}{V}\alpha B\] \[\frac{\Delta V}{V}\alpha \frac{1}{B}\] \[\frac{\Delta V}{V}\alpha {{B}^{2}}\]You need to login to perform this action.
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