Answer:
"2Ω" is the net resistance in the circuit.
Explanation:
The given resistors are:
R1 = 3Ω
R2 = 6Ω
The net resistance will be:
⇒ [tex]\frac{1}{R_{net}} =\frac{1}{R_1} +\frac{1}{R_2}[/tex]
On substituting the values, we get
⇒ [tex]\frac{1}{R_{net}} =\frac{1}{3} +\frac{1}{6}[/tex]
On taking L.C.M, we get
⇒ [tex]\frac{1}{R_{net}} =\frac{2+1}{6}[/tex]
⇒ [tex]\frac{1}{R_{net}} =\frac{3}{6}[/tex]
⇒ [tex]\frac{1}{R_{net}} =\frac{1}{2}[/tex]
On applying cross-multiplication, we get
⇒ [tex]R_{net}=2 \Omega[/tex]