Given R_1 = 1,Omega,C_1 = 2,mu F and R_2 = 2, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The time constant for an $RC$ circuit is given by $	au = R_{eq} C_{eq}$.For circuit $I$: $R_1$ and $R_2$ are in parallel,so $R_{eq} = \frac{R_1 R

Vedclass · Accepted Answer

$8/9, 18, 4$

Vedclass · Answer

(D) The time constant for an $RC$ circuit is given by $	au = R_{eq} C_{eq}$.For circuit $I$: $R_1$ and $R_2$ are in parallel,so $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{1 	imes 2}{1 + 2} = \frac{2}{3}\,\Omega$. $C_1$ and $C_2$ are in series,so $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{2 	imes 4}{2 + 4} = \frac{8}{6} = \frac{4}{3}\,\mu F$. Thus,$	au_I = \frac{2}{3} 	imes \frac{4}{3} = \frac{8}{9}\,\mu s$.For circuit $II$: $R_1$ and $R_2$ are in series,so $R_{eq} = R_1 + R_2 = 1 + 2 = 3\,\Omega$. $C_1$ and $C_2$ are in parallel,so $C_{eq} = C_1 + C_2 = 2 + 4 = 6\,\mu F$. Thus,$	au_{II} = 3 	imes 6 = 18\,\mu s$.For circuit $III$: $R_1$ and $R_2$ are in parallel,so $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{1 	imes 2}{1 + 2} = \frac{2}{3}\,\Omega$. $C_1$ and $C_2$ are in parallel,so $C_{eq} = C_1 + C_2 = 2 + 4 = 6\,\mu F$. Thus,$	au_{III} = \frac{2}{3} 	imes 6 = 4\,\mu s$.Therefore,the time constants are $8/9, 18, 4$.

Given $R_1 = 1\,\Omega$,$C_1 = 2\,\mu F$ and $R_2 = 2\,\Omega$,$C_2 = 4\,\mu F$. The time constants (in $\mu s$) for the circuits $I, II, III$ are respectively:

Similar Questions

In an $RC$ circuit while charging,the graph of $\log i$ versus time is as shown by the dotted line in the diagram,where $i$ is the current. When the value of the resistance is doubled,which of the solid curves best represents the variation of $\log i$ versus time?

The capacitance of $C_1$ and $C_2$ shown in the diagram is $C$. $C_1$ is initially uncharged and $C_2$ is given a charge $2C\varepsilon$ as shown. Which of the following graph represents charge on plate $B$ of capacitor $C_2$ as a function of time?

What will be the current through the $200 \Omega$ resistor in the given circuit, a long time after the switch $K$ is closed?

In the circuit shown in the figure,$R = \sqrt{\frac{L}{C}}$. Switch $S$ is closed at time $t = 0$. The current through $C$ and $L$ would be equal after a time $t$ equal to

Find the charge on the capacitor $C = 2 \, \mu F$ in the given circuit in $\mu C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module