The current in the circuit is 0.5 A..
Given data:
The voltage of the battery is V=12 V.
The resistances are,
[tex]\begin{gathered} R_1=4\text{ }\Omega \\ R_2=6\text{ \Omega} \\ R_3=8\text{ \Omega} \\ R_4=6\text{ }\Omega \end{gathered}[/tex]
The voltage of the battery is V=12 V.
The equivalent resistance of the given series connection of resistances can be calculated as,
[tex]\begin{gathered} R=R_1+R_2+R_3+R_4 \\ R=4+6+8+6 \\ R=24\text{ }\Omega \end{gathered}[/tex]
Applying the Ohm's law to calculate the current in the circuit as,
[tex]\begin{gathered} V=IR \\ (12)=I(24) \\ I=0.5\text{ A} \end{gathered}[/tex]
Thus, the current in the circuit is 0.5 A.
