A circuit consists of a battery with emf $𝐸$ and two resistors, $R=2\Omega$ and $R=4\Omega$, connected in series. If the emf of the battery is $12V$, what is the current through the battery?

• A. $2A$
• B. $3A$
• C. $2\frac{1}{3}A$
• D. $4A$

Answer: (C)

Verified step by step guidance

1. Identify that the resistors are connected in series, so the total resistance \(R_{total}\) is the sum of the individual resistances: \(R_{total} = R_1 + R_2\).
2. Substitute the given resistor values into the total resistance formula: \(R_{total} = 2\ \Omega + 4\ \Omega\).
3. Recall Ohm's Law, which relates voltage (\(V\)), current (\(I\)), and resistance (\(R\)) as \(V = IR\).
4. Rearrange Ohm's Law to solve for current: \(I = \frac{V}{R_{total}}\).
5. Substitute the emf of the battery (\$12\ V\() and the total resistance into the current formula to find the current through the battery: \)I = \frac{12\ V}{R_{total}}$.