Step 1: Understanding the Concept:
The circuit consists of a single closed loop containing two voltage sources (batteries) and one resistor.
We need to determine the net electromotive force (e.m.f.) driving the current and apply Ohm's law to find the current. Step 2: Key Formula or Approach:
Kirchhoff's Voltage Law (KVL) or the concept of net e.m.f. in a series circuit can be used.
If batteries are connected such that their positive terminals face each other (opposing polarities), the net e.m.f. is the difference of their voltages.
Current \( I \) is given by Ohm's Law:
\[ I = \frac{V_{\text{net}}}{R_{\text{total}}} \]
Step 3: Detailed Explanation:
Let's analyze the polarities of the batteries in the circuit diagram.
The top battery is 10 V, with its positive terminal on the left and negative on the right.
The bottom battery is 200 V, with its positive terminal on the left and negative on the right.
If we trace the loop in a clockwise direction:
- We go from the positive to the negative terminal of the 10 V battery, which is a potential drop (\( -10 \) V).
- We go through the 38 \(\Omega\) resistor, creating a potential drop (\( -I \times 38 \)).
- We go from the negative to the positive terminal of the 200 V battery, which is a potential gain (\( +200 \) V).
Alternatively, we can see that the two batteries are trying to push current in opposite directions around the loop.
The 200 V battery pushes current clockwise, while the 10 V battery pushes current counter-clockwise.
Since 200 V>10 V, the net current will flow in the clockwise direction.
The net e.m.f. (\( V_{\text{net}} \)) driving this current is the difference between the two voltages because they are opposing each other:
\[ V_{\text{net}} = 200\text{ V} - 10\text{ V} = 190\text{ V} \]
The total resistance (\( R_{\text{total}} \)) in the circuit is the single resistor:
\[ R_{\text{total}} = 38\ \Omega \]
Using Ohm's law, we can calculate the current \( I \):
\[ I = \frac{V_{\text{net}}}{R_{\text{total}}} = \frac{190}{38} \]
Performing the division:
\[ I = 5\text{ A} \]
The current flowing through the circuit is 5 amperes. Step 4: Final Answer:
The current flowing through the circuit is 5 A.
