Answer:
you know u should probably pay attention in class,
Since R1 and R2 are in parallel, their equivalent resistance can be calculated as:
1/R_eq = 1/R1 + 1/R2
1/R_eq = 1/4 + 1/4
1/R_eq = 1/2
R_eq = 2 Ω
The equivalent resistance of R1 and R2 is 2 Ω.
The total current flowing through the circuit is equal to the current flowing through R4, which is 0.8 A.
Using Ohm's Law, we can find the voltage drop across R4:
V4 = I4 * R4
V4 = 0.8 A * 8 Ω
V4 = 6.4 V
Since R3 and R4 are in series, the voltage drop across them is the same. Therefore, the voltage drop across R3 is also 6.4 V.
The total voltage drop across R1, R2, R3, and R4 is equal to the voltage of the battery, which is not given. However, we can find the voltage drop across R1 by subtracting the voltage drops across R2, R3, and R4 from the total voltage (V_tot):
V_tot = V1 + V2 + V3 + V4
Since R2, R3, and R4 are all in series, their equivalent resistance can be calculated as:
R_eq = R2 + R3 + R4
R_eq = 8 Ω + 8 Ω + 8 Ω
R_eq = 24 Ω
The current flowing through R2, R3, and R4 is equal to the total current, which is 0.8 A. Using Ohm's Law, we can find the voltage drop across these resistors:
V234 = I * R_eq
V234 = 0.8 A * 24 Ω
V234 = 19.2 V
Therefore, the voltage drop across R1 is:
V1 = V_tot - V234
V1 = ? - 19.2 V
We cannot determine the voltage drop across R1 without knowing the voltage of the battery or the total voltage in the circuit.