First, find the equivalent resistance: 1/R_eq = 1/6 + 1/3 = 1/2, R_eq = 2Ω. Then, I = V/R = 12V/2Ω = 6A.

Total resistance in parallel: 1/Rtotal = 1/6 + 1/3 = 1/2. Rtotal = 2Ω. Total current I = V/R = 12V/2Ω = 6A.

Total resistance in parallel: 1/Rtotal = 1/6 + 1/3 = 1/2. Rtotal = 2Ω. Total current I = V/R = 12V/2Ω = 6A.

Total current I = V / Req; Req = 1/(1/6 + 1/3) = 2Ω; I = 12V / 2Ω = 6A.

The total resistance in series is R_total = R1 + R2 = 4Ω + 6Ω = 10Ω.

Using the formula for resistors in series, R_total = R1 + R2, we have 12 = 4 + R2, thus R2 = 12 - 4 = 8 ohms.

Using the voltage divider rule, V1 = (R1 / (R1 + R2)) * Vtotal = (4 / (4 + 6)) * 12 = 4.8V.

Using the voltage divider rule, V1 = (R1 / (R1 + R2)) * Vtotal = (4 / (4 + 6)) * 12 = 4.8V.

According to Kirchhoff's current law, the current flowing into the junction must equal the current flowing out. Therefore, 10A - 6A = 4A is the current flowing through the junction.

According to Kirchhoff's Voltage Law, the net voltage in the loop is the sum of rises minus the sum of drops: 10V - 6V = 4V.

According to Kirchhoff's Voltage Law, the net voltage in the loop is the sum of voltage rises minus the sum of voltage drops: 20V - 15V = 5V.

The equivalent resistance for resistors in parallel is given by 1/R_eq = 1/R1 + 1/R2 + 1/R3. Thus, 1/R_eq = 1/2 + 1/3 + 1/6 = 1. Therefore, R_eq = 1Ω.

Using the formula for resistors in parallel, 1/R_eq = 1/R1 + 1/R2 + 1/R3, we find R_eq = 1 / (1/6 + 1/12 + 1/18) = 3 ohms.

1/R_eq = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/3 + 1/6 = 1. Therefore, R_eq = 3Ω.

The formula for equivalent resistance in parallel is 1/R_eq = 1/R1 + 1/R2 + 1/R3. Thus, 1/R_eq = 1/6 + 1/3 + 1/2 = 1/6 + 2/6 + 3/6 = 6/6 = 1, so R_eq = 1 ohm.

The equivalent resistance in parallel is given by 1/R_total = 1/R1 + 1/R2 + 1/R3. Thus, 1/R_total = 1/2 + 1/3 + 1/6 = 1Ω.

Using the formula for resistors in parallel, 1/R_eq = 1/R1 + 1/R2, we have 1/R_eq = 1/6 + 1/3 = 1/6 + 2/6 = 3/6. Therefore, R_eq = 6/3 = 2 ohms.

1/R_eq = 1/R1 + 1/R2 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, thus R_eq = 2Ω.

Total resistance R = 2Ω + 3Ω + 5Ω = 10Ω. Current I = V / R = 12V / 10Ω = 1.2A.

Total resistance R = R1 + R2 = 3Ω + 6Ω = 9Ω. Current I = V/R = 12V/9Ω = 4/3A.

Total resistance R = R1 + R2 = 4Ω + 8Ω = 12Ω; I = V / R = 12V / 12Ω = 1A.

According to KCL, I = total entering - total leaving = (2A + 3A) - 5A = 0A.

According to KCL, I = Total entering - Total leaving = (3A + 2A) - 5A = 0A.

Using voltage division, V_R1 = V_total * (R1 / (R1 + R2)) = 12V * (4Ω / (4Ω + 6Ω)) = 12V * (4/10) = 4.8V.

Using voltage division, V_R1 = (R1 / (R1 + R2)) * V_total = (4Ω / (4Ω + 6Ω)) * 12V = 4.8V.

According to Kirchhoff's voltage law, the sum of the voltage drops must equal the source voltage. Here, 4V + 5V = 9V, which is correct.

Using Ohm's law, I = V/R, we have I = 20V / 10 ohms = 2 A.

Using Ohm's law, I = V/R, we have I = 20V / 10 ohms = 2 A.

Using Ohm's Law, I = V / R = 36V / 12Ω = 3A.

Adding a resistor in parallel decreases the total resistance, which increases the total current according to Ohm's law.