Question

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

• A. 1 A from $C$ to $D$
• B. 0.2 A from $C$ to $D$
• C. 1.5 A from $D$ to $C$
• D. 1 A from $B$ to $A$ through $E$

Answer 1

Answer: (A)

1 A from C to D

Answer 2

Explanation of the solution:

1. Identify Circuit Type: The given circuit is a single-loop series circuit.
2. Calculate Net Electromotive Force (EMF):
   • The 10V battery has its positive terminal towards B and negative towards A, attempting to drive current from A to B.
   • The 5V battery has its positive terminal towards A and negative towards B, attempting to drive current from B to A.
   • Since the batteries oppose each other, the net EMF is the difference between their individual EMFs: $\text{Net EMF} = |10 \text{ V} - 5 \text{ V}| = 5 \text{ V}$
   • The direction of the net EMF is determined by the stronger battery, which is the 10V battery. Thus, the net EMF drives current from A to B (clockwise direction in the loop).
3. Calculate Total Equivalent Resistance:
   • All resistors in a series loop add up: \text{Total Resistance (R_{eq})} = 2 \, \Omega + 1 \, \Omega + 2 \, \Omega = 5 \, \Omega
4. Calculate Current Magnitude:
   • Using Ohm's Law (I = \text{V_{net}} / \text{R_{eq}}): $I = \frac{5 \text{ V}}{5 \, \Omega} = 1 \text{ A}$
5. Determine Current Direction:
   • As the net EMF drives current from A to B (clockwise), the current flows through the loop in the sequence A → B → C → D → A.
   • Therefore, the current flows from C to D through the 2 Ω resistor.