4 resistors

The resistors \(\displaystyle 70 \ \Omega\) and \(\displaystyle 30 \ \Omega\) are in parallel, so we can combine them as:

\(\displaystyle \left(\frac{1}{70} + \frac{1}{30}\right)^{-1} = 21 \ \Omega\)
 
The resistors \(\displaystyle 20 \ \Omega\) and \(\displaystyle 5 \ \Omega\) are in parallel, so we can combine them as:

\(\displaystyle \left(\frac{1}{20} + \frac{1}{5}\right)^{-1} = 4 \ \Omega\)
 
Now we can find how much current the source voltage provides.

Let \(\displaystyle V_s\) be the source voltage,

then,

\(\displaystyle V_s = IR\)

\(\displaystyle I = \frac{V_s}{R} = \frac{200}{21 + 4} = \frac{200}{25} = 8 \ \text{A}\)
 
The voltage across the resistors \(\displaystyle 70 \ \Omega\) and \(\displaystyle 30 \ \Omega\) is the same and it is:

\(\displaystyle V_{70 \ \Omega} = V_{30 \ \Omega} = 8 \times 21 = 168 \ \text{V}\)

The voltage across the resistors \(\displaystyle 20 \ \Omega\) and \(\displaystyle 5 \ \Omega\) is the same and it is:

\(\displaystyle V_{20 \ \Omega} = V_{5 \ \Omega} = 8 \times 4 = 32 \ \text{V}\)

Then,

\(\displaystyle V_{o} = 32 \ \text{V}\)
 
The current through the resistor \(\displaystyle R_{70 \ \Omega}\) is:

\(\displaystyle I = \frac{168}{70} = 2.4 \ \text{A}\)
 
The current through the resistor \(\displaystyle R_{20 \ \Omega}\) is:

\(\displaystyle I = \frac{32}{20} = 1.6 \ \text{A}\)
 
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