Answer
a). Suppose we have 3 5-ohms in series and this combination in parallel with a 5-ohm. Battery volt = 12V
Then , $\frac{1}{R_{eq}}=\frac{1}{5+5+5}+\frac{1}{5}=\frac{1}{15}+\frac{1}{5}$
Thus, $R_{eq}=3.75ohm$
So our assumption is correct.
b). Now current in the 3-5ohm resistors in series $=12/15=0.8A$
Current in the other 5-ohm resistor in parallel =$12/5=2.4A$
Voltage across the 3-5ohm resistors in series $=12/3=4V$
and voltage across the other 5ohm resistor in parallel $=12V$
Work Step by Step
a). Suppose we have 3 5-ohms in series and this combination in parallel with a 5-ohm. Battery volt = 12V
Then , $\frac{1}{R_{eq}}=\frac{1}{5+5+5}+\frac{1}{5}=\frac{1}{15}+\frac{1}{5}$
Thus, $R_{eq}=3.75ohm$
So our assumption is correct.
b). Now current in the 3-5ohm resistors in series $=12/15=0.8A$
Current in the other 5-ohm resistor in parallel =$12/5=2.4A$
Voltage across the 3-5ohm resistors in series $=12/3=4V$
and voltage across the other 5ohm resistor in parallel $=12V$
