The simplest calculation is to assume the worst: that at the instant of switch-on the cap on the far side of the bridge is not charged, and therefore presents almost a short circuit at the instant when power is first applied, for that first half-cycle of 50/60Hz, that's the worse case peak current you have to limit, the "inrush current" (to charge up the cap). Every cycle of the AC after that point, the cap will become more & more charged, drawing less & less current per cycle, until it settles to whatever power is being drawn by the circuit powered.
In that case, the max surge capability for that diode is 30 Amps for 8.3mS (i.e. 1 half-wave of the 60Hz mains frequency). This is straight off the datasheet.
In that case, R=V/I = 50 / 30 = 1.66 ohms. Let's say 1.8 or even 2.2 ohms just to be safe, and to choose a standard resistor value (E12-series).
The only problem with this approach is that resistor is always there, dissipating heat (diode is rated for 1.0A continuous, so presumably this is the design max current for the power supply, so Power = IIR = 1.0 * 1.0 * 1.8 = 1.8 Watts, so it won't be a small resistor! Other real-world solutions would use an NTC resistor whose resistance drops as it heats up, so it limits that inrush current, but then after a few seconds dissipates somewhat less heat (wasted)compared to the 'dumb' resistor. Or, a relay powered by the power-supply turns on after a short delay that shorts out the resistor.
I left out the units.
total resistance \$R\$
\$R_1\$ (between a and c) is in parallel to the rest of the resistors. The rest being \$R_2\$ parallel to \$R_3\$ (between a and b) in row to \$R_4\$ in parallel to \$R_5\$ (between b and c)
The general rule for two resistors in \$R_{a}\$ and \$R_{b}\$ in parallel:
$$R_a||R_b=\frac{R_aR_b}{R_a+R_b}$$
I gave it a try:
$$
\begin{align}
R & = R_1||(R_2||R_3 +R_4||R_5)\\
& = R_1||(R_{23} + R_{45})\\
& = \frac{R_1(R_{23} + R_{45})}{R_1+R_{23} + R_{45}}\\
R_{23} & = \frac{R_2R_3}{R_2+R_3} = \frac{20.46}{9.5}\\
R_{45} & = \frac{R_4R_5}{R_4+R_5} = \frac{56}{15.6}\\
R & = \frac{1(\frac{20.46}{9.5} + \frac{56}{15.6})}{1+\frac{20.46}{9.5} + \frac{56}{15.6}} = \frac{\frac{20.46 \times 15.6 + 56 \times 9.5}{15.6 \times 9.5}}{\frac{15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5}{15.6 \times 9.5}} =\frac{20.46 \times 15.6 + 56 \times 9.5}{15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5} \\
R & \approx 0.85
\end{align}
$$
voltage \$U_{ab}\$ via voltage divider
I use U for voltage, not V.
\$R_1\$ being in parallel to the rest of the resistors means that there's the same voltage over both of them. The voltage divider divides the voltage \$U_{ac}\$ into \$U_{ab} + U_{ac}\$
The general rule for a voltage divider for two resistors in \$R_{ab}\$ and \$R_{bc}\$ in row:
$$\frac{U_{ab}}{U_{ac}}=\frac{R_{ab}}{R_{ac}}=\frac{R_{ab}}{R_{ab} + R_{bc}}$$
I gave it a try:
$$
\begin{align}
\frac{U_{ab}}{U_{ac}} &= \frac{R_{23}}{R_{23} + R_{45}} \\
&= \frac{ \frac{20.46}{9.5}}{ \frac{20.46}{9.5} + \frac{56}{15.6}} = \frac{ \frac{20.46}{9.5}}{ \frac{20.46 \times 15.6 + 56 \times 9.5}{9.5 \times 15.6}} = \frac{20.46 \times 15.6}{20.46 \times 15.6 + 56 \times 9.5}\\
&\approx 0.3750 \\
U_{ac} &= 5 \\
U_{ab} &= \frac{20.46 \times 15.6}{20.46 \times 15.6 + 56 \times 9.5} \times 5\\
&\approx 1.8750
\end{align}
$$
current \$I_{R_2}\$ via current divider
The current divider divides the current \$I_{ab}\$ into \$I_{R_2} + I_{R_3}\$
The general rule for a current divider for two resistors in \$R_{a}\$ and \$R_{b}\$ in parallel:
$$\frac{I_{a}}{I_{a} + I_{b}}=\frac{I_{a}}{I_{ab}}=\frac{R_{a} || R_{b}}{R_{a}}= \frac{R_aR_b}{R_a(R_a +R_b)}=\frac{R_{b}}{R_{a} + R_{b}}$$
The general rule for resistance, voltage and current (ohm's law) :
$$R = \frac{U}{I} \iff I = \frac{U}{R}$$
I gave it a try:
$$
\begin{align}
I_{ab} &= \frac{U_{ab}}{R_{ab}} = \frac{U_{ab}}{R_{23}}\\
&= \frac{\frac{20.46 \times 15.6}{20.46 \times 15.6 + 56 \times 9.5} \times 5}{\frac{20.46}{9.5}} = \frac{15.6 \times 9.5}{20.46 \times 15.6 + 56 \times 9.5} \times 5\\
&\approx 0.8706\\
\frac{I_{R_2}}{I_{ab}} &= \frac{R_3}{R_2 + R_3} = \frac{3.3}{9.5} \\
&\approx 3.4747 \\
I_{R_2} &= \frac{R_3}{R_2 + R_3} \times I_{ab} = \frac{3.3}{9.5} \times \frac{15.6 \times 9.5}{20.46 \times 15.6 + 56 \times 9.5} \times 5 = \frac{15.6 \times 3.3}{20.46 \times 15.6 + 56 \times 9.5} \times 5\\
&\approx 0.3024\\
\end{align}
$$
power \$P_{R_1}\$
Given the overall voltage \$U_{ac}\$ and the resistance \$R_1\$ the power \$P_{R_1}\$ can be calculated.
The general rule for power:
$$P = U \times I$$
With ohm's law:
$$P =\frac{U^2}{R}$$
I gave it a try:
$$
\begin{align}
P_{R_1} &=\frac{U_{ac}^2}{R} =\frac{5^2}{\frac{20.46 \times 15.6 + 56 \times 9.5}{15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5}} =\frac{25 \times 15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5}{20.46 \times 15.6 + 56 \times 9.5} \\
&\approx 5.3528
\end{align}
$$
Best Answer
\$V_c\$ is exactly 0 since it is connected to GND which is defined as zero volts.
\$ V_a\$ is exactly 5 volts since it is connected to the positive end of a 5 volt battery whose negative terminal is connected to GND.
\$V_b\$ is at the junction of resistors connected across the battery so its voltage depends on the relative values of the resistors. The first circuit is a simple series connection. In accordance with Ohm's law, the current is equal to the voltage (5 volts) divided by the total series resistance \$(R1 + R2)\$. \$V_b\$ is equal to the voltage across \$R2\$ which is found by multiplying its resistance by the current. In this case, then, \$V_b\$ is equal to \$5/(R_1 +R_2)\$ times \$R_2\$ or \$5R_2/(R_1 +R_2)\$.
In the second circuit, the calculation is identical except that \$R_3\$ is now in parallel with \$R_2\$. The equivalent resistance of 2 parallel resistors is given by their product divided by their sum or \$R_2 R_3 / (R_2 + R_3) \$. \$V_b\$ in this circuit can then by found by substituting this equivalent resistance for \$R_2\$ in the previous equation.