The circuit in the image shows a combination of resistors and a 24V voltage source ($V_s = 24V$). The resistors are labeled:

• $R_1 = 6 \Omega$
• $R_2 = 8 \Omega$
• $R_3 = 4 \Omega$
• $R_4 = 12 \Omega$

It appears to be a bridge circuit, but it is drawn in a slightly non-standard form. To determine the current $I$ through the circuit, we can follow these steps:

Steps to Solve for Current $I$:

1. Simplify the circuit: We can combine resistors where possible.

   • $R_2$ and $R_3$ form a series combination because they are connected end-to-end. $R_{23} = R_2 + R_3 = 8 \Omega + 4 \Omega = 12 \Omega$
   • Now, $R_{23}$ and $R_4$ are in parallel, since they are connected across the same nodes. $\frac{1}{R_{eq}} = \frac{1}{R_{23}} + \frac{1}{R_4} = \frac{1}{12 \Omega} + \frac{1}{12 \Omega}$ Solving this, we get: $R_{eq} = 6 \Omega$

2. Total resistance in the circuit: Now that we have reduced the parallel part of the circuit, this resistance $R_{eq} = 6 \Omega$ is in series with $R_1$. Therefore, the total resistance $R_{total}$ is: $R_{total} = R_1 + R_{eq} = 6 \Omega + 6 \Omega = 12 \Omega$

3. Apply Ohm’s Law: Now, using Ohm's Law, we can find the current $I$: $I = \frac{V_s}{R_{total}} = \frac{24V}{12 \Omega} = 2A$

Thus, the current $I$ in the circuit is 2A.

Would you like further details on the process or have any other questions?

Related Questions:

1. How do we calculate the voltage drop across each resistor?
2. What is the power dissipated by each resistor in the circuit?
3. How would the current change if $R_4$ were increased to 20 $\Omega$?
4. How do we find the equivalent resistance in a more complex bridge circuit?
5. What is the significance of a balanced bridge circuit in practical applications?

Tip:

Always look for resistors in series or parallel to simplify complex circuits before applying Ohm's law for easier calculation.