Question:

A circuit consists of a 12V battery connected in series to three resistors. The resistors have resistances of 4 Ω, 6 Ω, and 8 Ω. Calculate:

a) The equivalent resistance of the circuit.

b) The total current flowing through the circuit.

c) The voltage drop across each resistor.

d) Power dissipated by each resistor.

Answer:

a) To calculate the equivalent resistance of the circuit, we need to find the total resistance when the resistors are connected in series. The formula to calculate the equivalent resistance in a series circuit is:

In this case, we have three resistors with resistances of 4 Ω, 6 Ω, and 8 Ω. Plugging these values into the formula:

b) To find the total current flowing through the circuit, we can use Ohm's Law. Ohm's Law states that the current in a circuit is equal to the voltage divided by the resistance. The formula to calculate the total current in a series circuit is:

Plugging in the values:

c) The voltage drop across each resistor in a series circuit is equal to the current flowing through the circuit multiplied by the resistance of that specific resistor. Using Ohm's Law, we can calculate the voltage drop across each resistor:

• Voltage drop across the 4 Ω resistor:

  $$V_1 = I_{total} \times R_1 = 0.67 A \times 4 \Omega = 2.67 V$$

• Voltage drop across the 6 Ω resistor:

  $$V_2 = I_{total} \times R_2 = 0.67 A \times 6 \Omega = 4 V$$

• Voltage drop across the 8 Ω resistor:

  $$V_3 = I_{total} \times R_3 = 0.67 A \times 8 \Omega = 5.33 V$$

d) The power dissipated by each resistor in a circuit can be calculated using the formula:

• Power dissipated by the 4 Ω resistor:

  $$P_1 = \frac{V_1^2}{R_1} = \frac{(2.67 V)^2}{4 \Omega} = 1.78 W$$

• Power dissipated by the 6 Ω resistor:

  $$P_2 = \frac{V_2^2}{R_2} = \frac{(4 V)^2}{6 \Omega} = 2.67 W$$

• Power dissipated by the 8 Ω resistor:

  $$P_3 = \frac{V_3^2}{R_3} = \frac{(5.33 V)^2}{8 \Omega} = 3.55 W$$