To find the voltage, first find the combined resistances of the resistors in parallel:

$\displaystyle \frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{2\Omega }+\frac{1}{3\Omega }$

$\displaystyle \frac{1}{R_T}=\frac{5}{6\Omega }$

$\displaystyle R_T=\frac{6}{5}\Omega$

Use Ohm's law to find the voltage.

$\displaystyle V=IR$

$\displaystyle V=30A\left(\frac{6}{5}\Omega\right)=36V$

Find the total resistance of the circuit, which can be determined using Ohm's law.

$\displaystyle V=IR$

$\displaystyle R_T=\frac{V}{I}$

$\displaystyle R_T=\frac{5}{1.25}\Omega=4\Omega$

Now, the resistance of the second resistor can be found. Since the two resistors are in parallel, they're related to the total resistance as follows:

$\displaystyle \frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}$

Rearrange and solve for $\displaystyle R_2$

$\displaystyle \frac{1}{R_2}=\frac{1}{R_T}-\frac{1}{R_1}$

$\displaystyle \frac{1}{R_2}=\frac{1}{4\Omega }-\frac{1}{6\Omega }=\frac{6-4}{24\Omega }=\frac{1}{12\Omega }$

$\displaystyle R_2=12\Omega$

Find the combined resistances for the resistors in parallel:

$\displaystyle \frac{1}{R_A}=\frac{1}{R_1}+\frac{1}{R_2}$

$\displaystyle \frac{1}{R_A}=\frac{1}{3\Omega}+\frac{1}{6\Omega}=\frac{9}{18\Omega}$

$\displaystyle R_A=2\Omega$

$\displaystyle \frac{1}{R_B}=\frac{1}{R_3}+\frac{1}{R_4}$

$\displaystyle \frac{1}{R_B}=\frac{1}{2\Omega}+\frac{1}{16\Omega}=\frac{18}{32\Omega}$

$\displaystyle R_B=\frac{16}{9}\Omega$

Combine these two combined series resistors to find the total resistance:

$\displaystyle R_T=R_A+R_B=2\Omega+\frac{16}{9}\Omega=\frac{34}{9}\Omega$

Find the total resistance of the circuit. First, calculate the values of the combined resistances of the resistors in parallel:

$\displaystyle \frac{1}{R_A}=\frac{1}{R_1}+\frac{1}{R_2}$

$\displaystyle \frac{1}{R_A}=\frac{1}{1\Omega}+\frac{1}{2\Omega}=\frac{3}{2\Omega}$

$\displaystyle R_A=\frac{2}{3}\Omega$

$\displaystyle \frac{1}{R_B}=\frac{1}{R_3}+\frac{1}{R_4}$

$\displaystyle \frac{1}{R_B}=\frac{1}{3\Omega}+\frac{1}{4\Omega}=\frac{7}{12\Omega}$

$\displaystyle R_B=\frac{12}{7}\Omega$

Therefore, the total resistance is:

$\displaystyle R_T=R_A+R_B=\frac{2}{3}\Omega+\frac{12}{7}\Omega=\frac{50}{21}\Omega$

Now, note that since $\displaystyle R_3$ and $\displaystyle R_4$ are in parallel, the voltage drop across them is the same. Use Ohm's law to relate current in terms of voltage and resistance.

$\displaystyle I=\frac{V_T}{R_T}$

Substitute into Ohm's law for the resistance across $\displaystyle R_3$:

$\displaystyle V_{3,4}=\frac{V}{R_T}R_B$

$\displaystyle V_{3,4}=10V\left(\frac{21}{50\Omega}\right)\left(\frac{12\Omega}{7}\right)$

$\displaystyle V_{3,4}=7.2V$

Find the total resistance of the circuit. First, calculate the values of the combined resistances of the resistors in parallel:

$\displaystyle \frac{1}{R_A}=\frac{1}{R_1}+\frac{1}{R_2}$

$\displaystyle \frac{1}{R_A}=\frac{1}{1\Omega}+\frac{1}{2\Omega}=\frac{3}{2\Omega}$

$\displaystyle R_A=\frac{2}{3}\Omega$

$\displaystyle \frac{1}{R_B}=\frac{1}{R_3}+\frac{1}{R_4}$

$\displaystyle \frac{1}{R_B}=\frac{1}{3\Omega}+\frac{1}{4\Omega}=\frac{7}{12\Omega}$

$\displaystyle R_B=\frac{12}{7}\Omega$

Therefore, the total resistance is:

$\displaystyle R_T=R_A+R_B=\frac{2}{3}\Omega+\frac{12}{7}\Omega=\frac{50}{21}\Omega$

From Ohm's law, we know that $\displaystyle \frac{V}{R_T}$ is the current traveling through the circuit.

$\displaystyle I=10V\left(\frac{21}{50\Omega}\right)=4.2A$

This current will be divided between $\displaystyle R3$ and $\displaystyle R4$, with more current taking the path of lower resistance. 

Total voltage drop across $\displaystyle R_{3,4}$:

$\displaystyle V_{3,4}=I_TR_{3,4}$

$\displaystyle V_{3,4}=4.2A\cdot\frac{12}{7}\Omega$

$\displaystyle V_{3,4}=7.2V$

$\displaystyle V_3=V_4=7.2V$

The current through $\displaystyle R_3$ is given by:

$\displaystyle I_3=\frac{V_3}{R_3}$

$\displaystyle I_3=\frac{7.2V}{3\Omega}=2.4A$

Begin by combining the resistors that are immediately in series:

$\displaystyle R_A=R_1+R_3=3\Omega+2\Omega=5\Omega$

$\displaystyle R_B=R_2+R_4=6\Omega+10\Omega=16\Omega$

Now to find the total resistance, combine these two new resistance values, which are in parallel:

$\displaystyle \frac{1}{R_T}=\frac{1}{5\Omega}+\frac{1}{16\Omega}=\frac{21}{80\Omega}$

$\displaystyle R_T=\frac{80}{21}\Omega$

To approach this problem, note that there are no other resistors (or combinations or resistors) beyond the parallel arrangement shown, so the voltage drop across the top $\displaystyle (R_1,R_3)$ and the bottom $\displaystyle (R_2,R_4)$ is the same and equal to the voltage across the circuit, $\displaystyle 10V$.

The voltage drop across $\displaystyle R_3$ can be found as:

$\displaystyle V_3=\frac{R_3}{R_1+R_3}V=\frac{3\Omega}{1\Omega+3\Omega}10V=\frac{3}{4}\cdot10V$

$\displaystyle V_3=7.5V$

To approach this problem, note that there are no other resistors (or combinations or resistors) beyond the parallel arrangement shown, so the voltage drop across the top $\displaystyle (R_1,R_3)$ and the bottom $\displaystyle (R_2, R_4)$ is the same and equal to the voltage across the circuit, $\displaystyle 10V$.

Furthermore, the current that passes through $\displaystyle R_1$ must be the same as the current that passes through $\displaystyle R_3$.

Therefore, the current that passes through them can be found by rearranging Ohm's law, solving for current.

$\displaystyle I_{1,3}=\frac{V}{R_1+R_3}=\frac{10V}{1\Omega+3\Omega}$

$\displaystyle I_{1,3}=2.5A$

The quickest way to approach this problem is to realize that the voltage drop across $\displaystyle Z_3$ is the same as the voltage drop across the combined resistances of $\displaystyle Z_1$ and $\displaystyle Z_2$. Since this parallel combination is the only presence of resistance in the circuit, this voltage drop must be the total voltage of the circuit, $\displaystyle 12V$.

Therefore, the current across $\displaystyle Z_3$ is:

$\displaystyle I_3=\frac{V_3}{Z_3}$

$\displaystyle I_3=\frac{12V}{16\Omega}$

$\displaystyle I_3=0.75A$

Realize that the voltage drop across the combined resistances of $\displaystyle Z_1$ and $\displaystyle Z_2$ must be equal to the voltage of the circuit, since the parallel combination is the only presence of resistance in the circuit. This voltage drop must be the total voltage of the circuit, $\displaystyle 12V$.

The current across $\displaystyle Z_1$ and $\displaystyle Z_2$ is the same, and is given as:

$\displaystyle I_1=I_2=\frac{12V}{Z_1+Z_2}$

$\displaystyle I_2=\frac{12V}{32\Omega}$

$\displaystyle I_2=0.375A$