Question

The magnitude and direction of the current in the following circuit is

• A. 1.5 A from B to A through E
• B. 0.2 A from B to A through E
• C. 0.5 A from A to B through E
• D. 5/9 A from A to B through E

Answer 1

The Correct Option is C

To determine the magnitude and direction of the current in the given circuit, we need to use Ohm's Law and the concept of series circuits.

Step-by-step Solution:

Identify the Total Resistance:

• The resistances in series: \(R_1 = 2 \, \Omega\), \(R_2 = 1 \, \Omega\), \(R_3 = 7 \, \Omega\).
• Total resistance \(R_{\text{total}} = R_1 + R_2 + R_3 = 2 + 1 + 7 = 10 \, \Omega\).

Calculate the Total Voltage:

• The batteries are opposing each other, so the net voltage is \(10 \, \text{V} - 5 \, \text{V} = 5 \, \text{V}\) from A to B.

Apply Ohm's Law to Find the Current:

• From Ohm's Law, \(I = \frac{V}{R}\).
• Substitute the values: \(I = \frac{5 \, \text{V}}{10 \, \Omega} = 0.5 \, \text{A}\).

Direction of Current:

• The current flows from positive to negative terminal, thus from A to B through E.

Conclusion: The magnitude of the current is \(0.5 \, \text{A}\) and it flows from A to B through E.