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(B) First,calculate the equivalent resistance $R_{eq}$ and equivalent capacitance $C_{eq}$ of the circuit.Two resistors of $2 \ M\Omega$ are in parall

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(B) First,calculate the equivalent resistance $R_{eq}$ and equivalent capacitance $C_{eq}$ of the circuit.Two resistors of $2 \ M\Omega$ are in parallel,so $R_{eq} = \frac{2 	imes 2}{2 + 2} \ M\Omega = 1 \ M\Omega = 10^6 \ \Omega$.Two capacitors of $2 \ \mu F$ are in parallel,so $C_{eq} = 2 + 2 = 4 \ \mu F = 4 	imes 10^{-6} \ F$.The time constant $	au = R_{eq} C_{eq} = (10^6 \ \Omega) 	imes (4 	imes 10^{-6} \ F) = 4 \ s$.The voltage across the capacitors at time $t$ is given by $V(t) = V_0(1 - e^{-t/	au})$.Given $V(t) = 4 \ V$,$V_0 = 10 \ V$,and $	au = 4 \ s$,we have:$4 = 10(1 - e^{-t/4})$$0.4 = 1 - e^{-t/4}$$e^{-t/4} = 0.6 = \frac{3}{5}$Taking the natural logarithm on both sides:$-t/4 = \ln(3/5) = \ln 3 - \ln 5$$-t/4 = 1.1 - 1.6 = -0.5$$t/4 = 0.5$$t = 2 \ s$.

At time $t=0$,a battery of $10 \ V$ is connected across points $A$ and $B$ in the given circuit. If the capacitors have no charge initially,at what time (in seconds) does the voltage across them become $4 \ V$? [Take $\ln 5=1.6, \ln 3=1.1$]

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