When two batteries are placed in opposition, the net driving voltage is the absolute difference of their emfs. Use the sign convention consistently to find the direction and magnitude of current.
Step 1: Understanding the Question:
The circuit consists of two voltage sources (+5 V and +3 V) and a 200 \(\Omega\) resistor. The “10 A” shown is likely a label for the ammeter (or a distractor); we ignore it for calculation. The batteries are probably connected in series opposing each other. We need the current in the circuit.
Step 2: Key Formula or Approach:
Ohm’s law: \(I = \frac{V_{\text{net}}}{R}\), where \(V_{\text{net}}\) is the net voltage when the batteries are connected in opposition.
Step 3: Detailed Explanation:
Assume the two batteries are connected so that their polarities oppose. The net voltage is the difference of their emfs:
\[
V_{\text{net}} = 5\ \text{V} - 3\ \text{V} = 2\ \text{V}.
\]
The only resistance given is \(R = 200\ \Omega\). Applying Ohm’s law:
\[
I = \frac{2\ \text{V}}{200\ \Omega} = 0.01\ \text{A} = 10^{-2}\ \text{A}.
\]
The “10 A” in the circuit statement does not affect the calculation; it might be a misprint or a red herring.
Step 4: Final Answer:
The current is \(10^{-2}\) A, which corresponds to option (D).
