Given:
The voltage, V=12.0 V
The resistances, R₁=200.0 Ω
R₂=600 Ω
To find:
The total current in the circuit.
Explanation:
In the figure, it can be seen that the resistors are in parallel to the battery and are in parallel to each other.
The equivalent resistance of the circuit is given by,
[tex]R_{eq}=\frac{R_1R_2}{R_1+R_2}[/tex]
On substituting the known values,
[tex]\begin{gathered} R_{eq}=\frac{200\times600}{200+600} \\ =150\text{ }\Omega \end{gathered}[/tex]
From Ohm's law,
[tex]V=IR_{eq}[/tex]
Where I is the total current in the circuit.
On substituting the known values,
[tex]\begin{gathered} 12.0=I\times150 \\ \Rightarrow I=\frac{12.0}{150} \\ =80.0\text{ mA} \end{gathered}[/tex]
Final answer:
The current I in the circuit is 80.0 mA
