Leaky current

Clearly, as the resistor and the capacitor form a loop, Kirchoff's Voltage Law can be applied. Let- q = charge on capacitor at any time 't', C = capacitance of the capacitor, i = current through the loop at time 't', R = resistance of the resistor. Hence, by Kirchoff's Voltage Law, $q/C + iR = 0 \rightarrow q/C = - iR \rightarrow - q/RC = dq/dt$ (as we know, i = dq/dt). Hence $- dt/RC = dq/q$ .

Integrating both sides, we get, $- t/RC = ln(q) + c$ . But q = VC, where V is potential drop across capacitor, so $- t/RC = ln(VC) + c$ . Now, when t=0, V=100V (given in the question). Hence, c = - ln(100C). Hence the equation becomes, $- t/RC = ln(V/100)$ . Now by the information in the question, V=10V when t=5s Hence, - 5/RC = ln(10/100) or -5/RC = ln(0.1) (1).

Now when another resistor is connected in parallel, the net resistance becomes R/2, and now initial potential across capacitor at this moment is 10V. Hence from above derivation, the equation becomes $- 2t/RC = ln(V/10)$ . Let's say it takes time 't' to drop to V=1V. So $- 2t/RC = ln(1/10) \rightarrow - 2t/RC = ln(0.1)$ (2).

Now dividing eqn(1) by eqn(2) $- 5/RC = - 2t/RC t = 5/2 =2.5$ seconds. Hence the result, that it would take 2.5 seconds to drop to 1V.

For the voltage to drop by a factor of $\frac 1{10}$ , it takes 5 seconds, at a rate of $\frac 1{RC}$ .

Now the resistance is changed to $\frac R2$ , so the rate is doubled.

With the same rate, it would again have taken 5 seconds for the voltage to drop by a factor of $\frac 1{10}$ , but with double rate, it will take half the time. Hence 2.5 seconds

Voltage across the capacitor plates at time $t$ in an RC circuit is given by

$V=V_0e^{\frac{-t}{RC}}$

Where $V_0$ is the initial voltage.

Initially $V_0=100\hspace{2mm}V$ . At $t=5\hspace{2mm}sec$ voltage $V=10\hspace{2mm}V$

$\Rightarrow 10=100e^{\frac{-5}{RC}}$

$\Rightarrow RC=\frac{5}{ln10}....(1)$

After the second resistor is put in parallel, the equivalent resistance $R_{eq} =\frac{R}{2}$

So, if $t$ is the time taken for the potential drop to equal $1\hspace{2mm}V$ ,

$1=10e^{\frac{-2t}{RC}}....(2)$

Putting the value of $RC$ from $(1)$ in $(2)$ and taking $ln$ ,

$t=\boxed{2.5 \hspace{2mm} sec}$

(i) At t=0, V=100. (ii) At t=5, V(5)=10 here R=R and C=C. (iii) AT t=? V(t)=1 here R=R/2 and C=C. we get the value of RC = 2.17147 by solving equation (ii). Substituting this in equation (iii) we get t=2.5

If it initally takes 5 seconds for the voltage drop to decrease by a factor of 10, then halving the resistance will halve the time to $\boxed{2.5}$ seconds.

In this case, we consider direct current circuit that obeys the Ohm's Law.

From the Ohm's Law, $V = IR = R\frac{dQ}{dt}$ . Since $C = \frac{Q}{V}$ , we can further transform this equation to $V = R\frac{d}{dt}CV = RCdV$ . Further manipulations give use the ODE $\frac{1}{V}dV = \frac{1}{RC}dt$ . Solving it gives us $\log{\frac{V_2}{V_1}} = \frac{t_{1,1}}{RC}$ , assuming $t_0 = 0$ .

Repeating the previous steps for the case with two resistors in parallel that have total resistance of $R/2$ , we get that $\log{\frac{V_2}{V_1}} = \frac{2t_{2,1}}{RC}$ .

We now notice that in cases with both a single resistor and resistors in parallel, $\log{\frac{V_2}{V_1}}$ is the same. From here, we can equate $\frac{t_{1,1}}{RC}$ and $\frac{t_{2,1}}{RC}$ , and it means that $\frac{2t_{2,1} - t_{1,1}}{RC} = 0$ . Therefore, $t_{2,1} = t_{1,1}/2 = \boxed{2.5}$ seconds.

It takes 5 seconds for the voltage across the capacitor to drop by a factor of 10 (from 100V to 10V) while one resistor is in the loop. When a second resistor is placed in parallel the total resistance doubles (1/Rt = 1/R + 1/R = 2/R, Rt=2R) and hence the time it takes for the voltage across the capacitor to decrease by a further factor of 10 is half the time, 2.5 seconds.

Since it is a discharging circuit, we can write the instantaneous voltage as v= $v_{o}$ $e^{-t/(RC)}$ where R is the resistance and C is the capacitance. At t=5seconds, 10=100 $e^{-5/(RC)}$ which yields RC= $\frac{5}{ln10}$ . After 5seconds R becomes $\frac{R}{2}$ as same resistance is connected in parallel. Now the instantaneous voltage becomes v= $v_{o}$ $e^{-2t/(RC)}$ where t is the time elapsed after 5 seconds. Let after t seconds, the voltage become 1V. Therefore, 1=10 $e^{-2t/(RC)}$ which yields t=2.5seconds.

Let the Resistance be R and Capacitance be C . We know that for a discharging capacitor:

$V=V_{0}*e^{-t/RC}$

We plug in $V=10, V_{0}=100 , t=5$ , take log of both sides and simplify to get $RC=5/ln10$

Next when the identical resistance is connected in parallel with the original resistance R, the equivalent resistance becomes R/2 (When n identical resistances R are connected in parallel, the equivalent resistance is R/n ).

For the next case, we plug in $V=1,V_{0}=10, R=R/2$ in the same equation:

$V=V_{0}*e^{-t/RC}$

Then we use $RC=5/ln10$ , take log, simplify to obtain $t=2.5$

Note : There's also a simpler, shorter and more logical way *Method 2: * We analyze the equation:

$V=V_{0}*e^{-t/RC}$

Taking log of both sides and solving for t, we obtain::

$t=-RC*ln(V/V_{0})$

We observer that in both the cases in the problem, (\V/V_{0}) is the same and equal to 0.1. Also, C is same for both cases. The only differing case is the value of R , which gets halved in the next case. Since everything else is constant, t is directly proportional to R . So, as R gets halved, So does t . So, t=2.5