In the simple circuit shown in fig. I. \(F\) is a \(24 V\) battery. Calculate the current I
Explanation
The equivalent resistance for the \(12 \Omega\) and \(12 \Omega\) resistors in parallel is given by
$$
\begin{aligned}
R &=\frac{\text { product }}{\text { sum }}=\frac{12 \times 12}{12+12} \\
&=\frac{144}{24}=6 \Omega
\end{aligned}
$$
The \(6 \Omega\) resistors is now in series with the \(6 \Omega\) resistors, giving a combined resistance of
$$
\begin{aligned}
&6 \Omega+6 \Omega=12 \Omega, \\
&I=\frac{V}{R_{e q}}=\frac{24}{12}=2.0 A
\end{aligned}
$$
