Answer
Final answers
Circuit a)
$I_1=0.3\ A\\I_2=-0.3\ A\\V_1=-6\ V\\$
Circuit b)
$V_1=5\ V\\I_2=-0.167\ A\\$
Circuit c)
$I_1=-1.25\ A\\I_2=0.75\ A\\V_1=52.5\ V$
Work Step by Step
Circuit a)
- Terminal of $V_1$ is open circuited; this means that no current flows across it
- Thus $I_1=-I_2$
Apply KVL
$-15-I_1\times 30+I_2\times 20=0$
$-15-I_1\times 30-I_1\times 20=0$
$-15-I_1\times(30+20)=0$
$I_1=\frac{15}{30+20}=\frac{15}{50}=\frac{3}{10}=0.3\ A$
Find $I_2$ from the following relation
$I_2=-I_1$
$I_2=-0.3\ A$
$V_1$ is parallel to resistor $20\Omega$. By applying KVL
$V_1=I_2\times 20=-0.3\times 20=-6\ V$
Circuit b)
Apply KVL
$-V_1+\frac{1}{4}\times 20=0$
$V_1=\frac{1}{4}\times 20=5\ V$
Again, apply KVL
$-V_1-I_2\times 30=0$
$V_1=-I_2\times 30$
$I_2=\frac{V_1}{30}=\frac{5}{30}=\frac{1}{6}=-0.167\ A$
circuit c)
Apply KVL
$-I_2\times 20 - (-0.5)\times 30=0$
$I_2=\frac{0.5\times 30}{20}=\frac{3}{4}=0.75\ A$
Apply KCL
$I_1+I_2-(-0.5)=0$
$I_1+I_2+0.5=0$
$I_1=-(I_2+0.5)$
$I_1=-(0.75+0.5)$
$I_1=-1.25\ A$
Apply KVL
$-V_1-(-1.25)\times 30+0.75\times 20=0$
$-V_1+1.25\times 30+0.75\times 20=0$
$V_1=0.75\times 20+1.25\times 30=52.5 V$
