3. For the circuit shown below, find the current through and the potential difference across cach resistor. Place your results in a table for case of reading. 12 V 40 ww 12 02 24 02 ww ww 3Ω 50 ww 3V:

Here $4 \Omega$ and $12 \Omega$ are parallel in the circuit

$\frac{4\times 12}{4+12}=3 \Omega$

Now $3 \Omega$and $5 \Omega$ are in series

$3\Omega +5\Omega =8 \Omega$

Then $8 \Omega$ and $24 \Omega$ are parallel in the circuit

$\frac{8\times 24}{8+24}=6 \Omega$

Then  $6 \Omega$and $1 \Omega$ are in series

$6+1=7 \Omega$

Total current in the circuit

$I=\frac{12-3}{7}=1.285 A$

Current through $3 \Omega$ is $1.285 A$

The voltage across $3 \Omega is$

$V_{3\Omega }=IR=1.285\times 3=3.855 V$

Current through $24 \Omega$ is

$\frac{1.285\times 8}{24+8}=0.3212 A$

Voltage across $24 \Omega$ is 

$V_{24\Omega }=0.3212\times 24=7.71 V$

Current through $8\Omega$ is 

$1.285\times \frac{24}{24+8}=0.963 A$

current through $5 \Omega is$ $0.963 A$

Voltage across $5 \Omega$ is 

$V_{5 \Omega }=0.963\times 5=4.815 V$

current through $4 \Omega is 0.963\times \frac{12}{16}=0.722 A$ 

current through $12 \Omega is 0.963\times \frac{4}{16}=0.2407 A$

Voltage across $4 \Omega$ is 

$V_{4\Omega }=0.722\times 4=2.888 V$

Voltage across $12 \Omega$ is 

$V_{12\Omega }=0.2407\times 12=2.884 V$