The potential difference is found from \( \Delta V = Q/C = 1 \mu \text{C} / 1 \mu \text{F} = 1 \text{V} \).

The potential difference must be the same as for \( V_C - V_D \). When there is no current, there is no potetial drop across the resistance, hence the potential in \( A \) and \( C \) must be the same and the potential in \( B \) and \( D \) is the same.

Kirchoff's law for potentials around the circuit gives that: $$ \begin{equation} V_0 - R_1 I - \Delta V = 0 \; , \tag{8.1} \end{equation} $$ where \( V_0 \) is the emf of the battery and \( \Delta V \) is the voltage drop across the capacitor. We insert the values and find that $$ \begin{equation} V_0 - \Delta V = 9 \text{V} - 1 \text{V} = 8 \text{V} = I R_1 \; \Rightarrow \; I = 8 \text{V}/(100 \text{k} \Omega) = 8 V/(10^5 \Omega) = 8 \times 10^{-5} \text{A} = 80 \mu \text{A} \; . \tag{8.2} \end{equation} $$

After a long time, when the current is zero, Kirchoff's law for potentials around the circuit gives that: $$ \begin{equation} V_0 - \Delta V = 0 \, \Rightarrow \, \Delta V = V_0 \; , \tag{8.3} \end{equation} $$ where \( I = 0 \) and hence the potential drop over \( R_1 \) is \( I R_1 = 0 \). We find the charge from \( Q = CV \), giving $$ \begin{equation} Q = C V = 1 \mu \text{F} \, 9 \text{V} = 9 \mu \text{C} \; . \tag{8.4} \end{equation} $$ Since we started with \( Q_0 = 1 \mu \text{C} \), we see that the amount of charge that has flowed through \( R_1 \) must be \( 8 \mu \text{C} \).

We apply Kirchoff's law for potentials in this case, in the opposite direction as before (counter-clockwise). In this case, the potential increases across the capacitor and drops over each of the resistors: $$ \begin{equation} \Delta V - I R_1 - I R_2 = \Delta V - I (R_1 + R_2) = 0 \, \Rightarrow \, I = \frac{\Delta V}{R_1 + R_2} = \frac{9 \text{V}}{150 \text{k} \Omega} = 60 \mu \text{A} \; . \tag{8.5} \end{equation} $$

The current will decrease. The current will reduce the charge across the capacitor, which will reduce the potential difference across the capacitor, which in turn will reduce the current. The current will approach zero as time approaches infinity.

We have found this to be \( R = 1/(4 \pi \sigma) (1/a - 1/b) \)

We have found this to be \( C = 4 \pi \epsilon /(1/a - 1/b) \).

Some current will leak through the conducting part of the there sphere, so the sphere will act both as a resistor and as a capacitor. The inner sphere is the inner side of both the resistance and the capacitance of the system. They will therefore have the same potential, which means that they are connected in parallel.

Immediately after we connect the battery there will be no charge on the capacitor so that the voltage drop across the capacitor is zero. All the current will therefore flow into the capacitor, which has zero resistance. The current will therefore flow through the battery, through the resistor \( r \) and through the capacitor that has zero potential drop. Kirchoff's voltage law gives us \( V_0 - I r - 0 = 0 \) and \( I = V/r \).

After a very long time the capacitor will be fully charged and there will be no more current flowing into the capacitor. All the current will therefore flow through the resistor \( R \). Kirchoff's voltage law for a circuit through the battery, resistor \( r \) and resistor \( R \) gives \( V_0 - I r - I R = 0 \) and \( I = V_0/(r + R) \).

We expect it to fall off from \( V_0/r \) to \( V_0/(r + R) \) with an exponential time dependence as illustrated in the figure.

We use Kirchoff's current law on the junction into the the \( R \) and \( C \) parts of the circuit, where \( -I + I_R + I_C = 0 \) or \( I = I_R+I_C \).

Kirchoff's voltage law on a circuit through the resistor \( R \) and the capacitor \( C \). This gives us: \( I_R R - V_C = 0 \), where \( V_C = Q_C/C \), so that \( I_R R = Q_C/C \). We take the time derivative on both sides to find \( I_C \): $$ \begin{equation} \frac{\d I_R}{\d t}R = \frac{1}{C}\frac{\d Q_C}{\d t} = \frac{1}{C} I_C \; . \tag{8.7} \end{equation} $$ We insert that \( I_C = I - I_R \), getting: $$ \begin{equation} \frac{\d I_R}{\d t}R = \frac{1}{C} \left( I - I_R \right) \; . \tag{8.8} \end{equation} $$

Kirchoff's voltage law for a part through the battery and both resistors gives: \( V_0 - I r - I_R R = 0 \), which gives that $$ \begin{equation} I = \frac{V_0 - I_R R}{r} \; . \tag{8.10} \end{equation} $$

We insert $$ \begin{equation} I = \frac{V_0 - I_R R}{r} \; . \tag{8.12} \end{equation} $$ into $$ \begin{equation} \frac{\d I_R}{\d t}R = \frac{1}{C} \left( I - I_R \right) \; . \tag{8.13} \end{equation} $$ getting $$ \begin{equation} RC\frac{\d I_R}{\d t} = \frac{V_0}{r} - I_R \frac{R}{r} - I_R = \frac{V_0}{r} - \left( 1 + \frac{R}{r} \right) I_R \; . \tag{8.14} \end{equation} $$

In the limit when \( t \rightarrow \infty \) we expect \( \d I_R / \d t= 0 \) and $$ \begin{equation} I_R = \frac{\frac{V_0}{r}}{ 1 + \frac{R}{r}} = \frac{V_}{r + R} \; , \tag{8.15} \end{equation} $$ where we also realize that in this limit there is no current through the capacitor so that \( I = I_R \). This corresponds to what we found above.

\( I = \frac{R_1+R_2}{R_1R_2}V_0 \).

The simplest way to solve this problem, is to compute the equivalent resistance and then plug it into Ohm's law. We've got a parallel circuit on our hands, so the equivalent resistance is given by $$ \begin{equation} R = \left(\frac{1}{R_1} + \frac{1}{R_2}\right)^{-1} = \left(\frac{R_1 + R_2}{R_1R_2}\right)^{-1} = \frac{R_1R_2}{R_1+R_2}. \tag{8.16} \end{equation} $$ Ohm's law then yield $$ \begin{equation} I = \frac{V_0}{R} = \frac{R_1+R_2}{R_1R_2}V_0, \tag{8.17} \end{equation} $$ which is what we set out to find.

\( P = \frac{R_1+R_2}{R_1 R_2}V_0^2 \).

The power of the current is given by the usual formula \( P = VI \), so in our case the power delivered by the battery is $$ \begin{equation} P = V_0I = \frac{R_1+R_2}{R_1 R_2}V_0^2. \tag{8.18} \end{equation} $$

\( V_a-V_b = \left(\frac{R_1}{R_1+R_3} - \frac{R_2}{R_2+R_x}\right)V_s = \frac{R_1R_x - R_2R_3}{(R_1+R_3)(R_2+R_x)}V_s. \)

Let's denote the first node in the wheatstone bridge as \( c \) (that is, the node where the current enters the parallel part of the circuit). Let \( V_G \equiv V_a - V_b \). We assume that the resistance of the voltmeter is infinite such that no current flows through it. Since it's a parallel circuit, the voltage across \( R_1 \) and \( R_3 \) is the same as the voltage across \( R_2 \) and \( R_x \). Since \( R_1 \) and \( R_3 \) are connected in series, the same current flows through both resistors, which we'll denote as \( I_{1,3} \). The same goes for the current through \( R_2 \) and \( R_x \) which will be denoted \( I_{2,x} \).

First we need the currents flowing through the parallel circuit. These are obtained through Ohm's law: $$ \begin{equation} I_{1,3} = \frac{V_s}{R_1+R_3} \; , \qquad \text{and} \qquad I_{2,x} = \frac{V_s}{R_2+R_x} \; , \tag{8.19} \end{equation} $$ Now, the voltage difference across \( R_1 \) is $$ \begin{equation} V_a - V_c = R_1 I_{1,3} = \frac{R_1}{R_1+R_3}V_s \; . \tag{8.20} \end{equation} $$ Similarly, the voltage difference across \( R_2 \) is given by $$ \begin{equation} V_b - V_c = R_2 I_{2,x} = \frac{R_2}{R_2+R_x}V_s \; . \tag{8.21} \end{equation} $$ Now we can obtain \( V_G \) by the following manipulation: $$ \begin{align} V_G & = & V_a-V_b = (V_a - V_c) - (V_b-V_c) \tag{8.22}\\ & = & \left(\frac{R_1}{R_1+R_3} - \frac{R_2}{R_2+R_x}\right)V_s = \frac{R_1R_x - R_2R_3}{(R_1+R_3)(R_2+R_x)}V_s \; . \tag{8.23} \end{align} $$

\( R_x = \frac{R_2R_3}{R_1} \)

Since \( R_3 \) is adjusted until \( V_G = V_a-V_b = 0 \), we get the equation $$ \begin{equation} V_G = \frac{R_1R_x - R_2R_3}{(R_1+R_3)(R_2+R_x)}V_s = 0 \; , \tag{8.24} \end{equation} $$ which is only zero if \( R_1R_x - R_2R_3 = 0 \), meaning $$ \begin{equation} R_x = \frac{R_2R_3}{R_1} \; . \tag{8.25} \end{equation} $$

\( Q(0) = 0 \), \( V(0) = 0 \).

Immediately before the circuit is closed \( Q=0 \). Immediately after the circuit is closed, the charge is the same. The potential drop over the capacitor is then \( V = Q/C = 0 \).

\( V = V_0 \), \( Q = C V_0 \)

After a long time, the current approaches zero and the potential fall over the capacitor is the same as the potential increase over the battery, \( V = V_0 \). The charge on the capacitor is then \( Q = C V_0 \).

\( V(t) = V_0 ( 1-\exp(-t/(R_0 C))) \)

Kirchoff's law for the circuit gives that \( V_0 - I R_0 - V(t) = V_0 - R_0 dQ/dt - Q/C = 0 \), where \( Q(0) = 0 \). We rewrite this equation as \( dQ/dt = -1/(R_0 C) Q + V_0/R_0 \). We first solve the homogeneous equation: \( dQ/dt = -Q/(R_0 C) \), which has solutions \( Q(t) = Q_0 \exp(-t/\tau) \), where \( \tau = R_0 C \). We use this as a basis for the full solution, \( Q(t) = A + B \exp(-t/\tau) \). We know that \( Q(0) = 0 = A + B \), which gives \( A = -B \) and \( Q(t) = A ( 1 - \exp (-t/\tau)) \). Finally, we use that after a long time \( Q(t) = C V_0 \) hence \( A = C/V_0 \) and \( Q(t) = (C V_0) ( 1 - \exp(-t/\tau)) \). We use this to find the potential, \( V(t) = Q/C = V_0 (1 - \exp(-t/\tau)) \).