Question:

Consider a circuit with two resistors connected in parallel, as shown in the figure below. The resistors have resistances R1 = 4 Ω and R2 = 6 Ω, and are connected to a 12 V battery. The circuit is closed and a steady current is flowing through it.

a) Calculate the total resistance of the circuit.

b) Determine the total current flowing through the circuit.

c) Calculate the current flowing through resistor R1.

d) Calculate the current flowing through resistor R2.

e) Determine the power dissipated in resistor R1.

f) Determine the power dissipated in resistor R2.

g) Calculate the total power supplied by the battery.

Assume all the resistors are ohmic and ignore any internal resistance of the battery.

Answer:

a) The total resistance of the parallel circuit can be calculated using the formula:

1/R_total = 1/R1 + 1/R2

Substituting the given values:

1/R_total = 1/4 + 1/6

1/R_total = 3/12 + 2/12

1/R_total = 5/12

Therefore, R_total = 12/5 Ω.

b) The total current flowing through the circuit can be calculated using Ohm's Law:

V = I_total * R_total

12 V = I_total * (12/5 Ω)

I_total = (12 V) / (12/5 Ω)

I_total = 5 A

Therefore, the total current flowing through the circuit is 5 A.

c) In a parallel circuit, the current through each resistor is the same as the total current. Therefore, the current flowing through resistor R1 is also 5 A.

d) Similarly, the current flowing through resistor R2 is also 5 A.

e) The power dissipated in a resistor can be calculated using the formula:

P = I^2 * R

Substituting the given values for resistor R1:

P_R1 = (5 A)^2 * 4 Ω

P_R1 = 100 W

Therefore, the power dissipated in resistor R1 is 100 W.

f) Similarly, substituting the given values for resistor R2:

P_R2 = (5 A)^2 * 6 Ω

P_R2 = 150 W

Therefore, the power dissipated in resistor R2 is 150 W.

g) The total power supplied by the battery can be calculated using the formula:

P_total = V * I_total

P_total = 12 V * 5 A

P_total = 60 W

Therefore, the total power supplied by the battery is 60 W.