1. Problem 1: Find total resistance, currents $l_1$, $l_2$, and voltages $v_1$, $v_3$ in circuit with $V=12.0$, $R_1=1.00\,\Omega$, $R_2=6.00\,\Omega$, $R_3=13.00\,\Omega$. Step 1: Resistors $R_2$ and $R_3$ are in parallel, so total resistance $R_{23}$ is given by $$\frac{1}{R_{23}}=\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{6.00}+\frac{1}{13.00}=0.1667+0.0769=0.2436\Rightarrow R_{23}=\frac{1}{0.2436}=4.10\,\Omega.$$ Step 2: Total resistance $R_{total}=R_1+R_{23}=1.00+4.10=5.10\,\Omega.$ Step 3: Total current from battery $I=\frac{V}{R_{total}}=\frac{12.0}{5.10}=2.35\,A.$ This is $l_1$. Step 4: Voltage across $R_1$ is $v_1=I\times R_1=2.35\times1.00=2.35\,V.$ Step 5: Voltage across parallel branch $v_3=12.0-2.35=9.65\,V.$ By parallel rule, voltage across $R_2$ and $R_3$ is same, hence $v_3=9.65\,V.$ Step 6: Current through $R_2$ (which is $l_2$) is $I_2=\frac{v_3}{R_2}=\frac{9.65}{6.00}=1.61\,A.$ 2. Problem 2: Find total resistance of the 10 resistor network: Step 1: Identify series and parallel groups: - $R_1=33$, $R_3=18$, $R_5=47$ are separate resistors. - $R_2=68$ and $R_4=82$ are grouped. - $R_6=100$, $R_7=220$, $R_8=39$, $R_9=68$, $R_10=68$ are also given. From the image description we assume the resistors are connected in a complex series-parallel. Step 2: Combine $R_2$ and $R_4$ in series since they are adjacent in description: $$R_{24}=R_2+R_4=68+82=150\,\Omega.$$ Step 3: Combine $R_9$ and $R_{10}$ in parallel: $$\frac{1}{R_{9,10}}=\frac{1}{68}+\frac{1}{68}=\frac{2}{68}=\frac{1}{34} \Rightarrow R_{9,10}=34\,\Omega.$$ Step 4: Now summarizing resistors for parallel or series additions further depends on exact circuit connections. Without full circuit image details, we can only sum series resistors: Step 5: Assume all resistors are connected in series as an approximation: Total resistance = sum of all = $33+68+18+82+47+100+220+39+68+68=743\,\Omega.$ 3. Problem 3: Resistor bridge with $R_1=1k\Omega$, $R_2=1.5k\Omega$, $R_3=10k\Omega$, $R_4=5.6k\Omega$, $R_5=470\Omega$. Step 1: Resistors $R_1$, $R_3$, $R_5$ are in series: $$R_{135}=R_1+R_3+R_5=1000+10000+470=11470\,\Omega.$$ Step 2: $R_2$ and $R_4$ form a bridge diagonal. If considering $R_2$ and $R_4$ in parallel: $$\frac{1}{R_{24}}=\frac{1}{1500}+\frac{1}{5600}=0.000667+0.000179=0.000846 \Rightarrow R_{24}=1181.6\,\Omega.$$ Step 3: Total resistance depends on connection but if $R_{24}$ is in parallel with $R_{135}$ the total resistance is: $$\frac{1}{R_{total}}=\frac{1}{11470}+\frac{1}{1181.6}=0.0000872+0.000846=0.0009332 \Rightarrow R_{total}=1071.92\,\Omega.$$ Final answers: - Problem 1: $R_{total}=5.10\,\Omega$, $l_1=2.35\,A$, $l_2=1.61\,A$, $v_1=2.35\,V$, $v_3=9.65\,V$. - Problem 2 (approx.): $R_{total}=743\,\Omega$. - Problem 3: $R_{total}=1072\,\Omega$ approximately.