What does this electronic symbol represent?

Step 1: Identify the circuit configuration from the diagram. It appears to be a series-parallel circuit with a voltage source at the bottom (12 V battery), two 4 Ω resistors in parallel on the upper branch, and a 6 Ω resistor in series with that combination.

Step 2: Calculate the equivalent resistance of the parallel branch.

Formula: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$

Substitute: $\frac{1}{R_p} = \frac{1}{4 \, \Omega} + \frac{1}{4 \, \Omega} = \frac{1}{2 \, \Omega^{-1}}$

Simplify: $R_p = 2 \, \Omega$

Step 3: Add the series resistor for total resistance.

Formula: $R_total = R_p + R_s$

Substitute: $R_total = 2 \, \Omega + 6 \, \Omega = 8 \, \Omega$

Step 4: Calculate total current from the battery.

Formula: $I_total = \frac{V}{R_total}$

Substitute: $I_total = \frac{12 \, V}{8 \, \Omega} = 1.5 \, A$

Step 5: Voltage drop across series resistor.

Formula: $V_s = I_total \times R_s$

Substitute: $V_s = 1.5 \, A \times 6 \, \Omega = 9 \, V$

Step 6: Voltage drop across parallel branch.

$V_p = 12 \, V - 9 \, V = 3 \, V$

Step 7: Current through each parallel resistor.

Formula: $I = \frac{V_p}{R}$

Substitute for each: $I_1 = I_2 = \frac{3 \, V}{4 \, \Omega} = 0.75 \, A$

Total current = 1.5 A, V across 6 Ω = 9 V, I through each 4 Ω = 0.75 A

1.5 A