For RC charging circuit as shown in figure the switch is closed at t=0 It is found that at t=t1 the charge on capacitor is CE2 (at t=0 the capacitor is uncharged while t2 is the time taken for the charge on the capacitor to increase from CE2 to CE, then mark the correct statements.

For RC charging circuit., charge on capacitor as a function of time is given by q=CE1−e−t/RC It is given that q=CE2 at t=t1 , Substjtuting these values in above equation we get CE2=CE1−et1RC⇒t1=RCln⁡2It is a known fact that initially charging Is fast and then charging slows down exponentially and complete charging will take infinite time. Hence t1<t2.

For RC charging circuit as shown in figure the switch is closed at t=0 It is found that at t=t1 the charge on capacitor is CE2 (at t=0 the capacitor is uncharged while t2 is the time taken for the charge on the capacitor to increase from CE2 to CE, then mark the correct statements.

For RC charging circuit., charge on capacitor as a function of time is given by q=CE1−e−t/RC It is given that q=CE2 at t=t1 , Substjtuting these values in above equation we get CE2=CE1−et1RC⇒t1=RCln⁡2It is a known fact that initially charging Is fast and then charging slows down exponentially and complete charging will take infinite time. Hence t1<t2.

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For $R C$ charging circuit as shown in figure the switch is closed at $t = 0$ It is found that at $t = t_{1}$ the charge on capacitor is $\frac{CE}{2} (at t = 0$ the capacitor is uncharged while $t_{2}$ is the time taken for the charge on the capacitor to increase from $\frac{C E}{2} to CE,$ then mark the correct statements.

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$t_{2} > t_{1}$

$t_{1} = R C \times \ln 2$

$t_{2}$ is very large as compared to $t_{1}$

$t_{2} = R C in 2$

answer is A, B, C.

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Detailed Solution

For RC charging circuit., charge on capacitor as a function of time is given by $q = C E [1 - e^{- t / R C}]$ It is given that $q = \frac{C E}{2} at t = t_{1}$ , Substjtuting these values in above equation we get

$\frac{C E}{2} = C E [1 - e^{\frac{t_{1}}{R C}}] \Rightarrow t_{1} = R C \ln 2$

It is a known fact that initially charging Is fast and then charging slows down exponentially and complete charging will take infinite time. Hence $t_{1} < t_{2}$ .

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