Question:

A circuit consists of a battery with an emf of 9.0 V and internal resistance of 0.5 Ω, connected to three resistors, R1 = 2.0 Ω, R2 = 4.0 Ω, and R3 = 6.0 Ω, as shown in the diagram below:

(a) Calculate the total resistance of the circuit.

(b) Calculate the current flowing through the circuit.

(c) Calculate the potential difference across each resistor.

(d) Calculate the power dissipated by each resistor.

Answer:

(a) The total resistance of the circuit, RT, can be calculated by summing the individual resistances:

RT = R1 + R2 + R3

= 2.0 Ω + 4.0 Ω + 6.0 Ω

= 12.0 Ω

Answer:

The total resistance of the circuit is 12.0 Ω.

(b) Using Ohm's Law, we can calculate the current flowing through the circuit using the equation I = ΔV / R, where ΔV is the potential difference across the circuit and R is the total resistance.

The potential difference across the circuit, ΔV, can be calculated by subtracting the potential drop across the internal resistance of the battery from the battery's emf:

ΔV = emf - (internal resistance * I)

= 9.0 V - (0.5 Ω * I)

We can substitute this value of ΔV in the Ohm's Law equation and solve for I:

I = ΔV / RT

I = [9.0 V - (0.5 Ω * I)] / 12.0 Ω

Multiplying both sides of the equation by 12.0 Ω to eliminate the denominator, we get:

12.0 Ω * I = 9.0 V - (0.5 Ω * I)

12.0 Ω * I + 0.5 Ω * I = 9.0 V

12.5 Ω * I = 9.0 V

I = 9.0 V / 12.5 Ω

Answer:

The current flowing through the circuit is approximately 0.72 A.

(c) To calculate the potential difference across each resistor, we can use Ohm's Law. The potential difference across each resistor can be calculated by using the equation ΔV = I * R, where I is the current flowing through the circuit and R is the resistance of the resistor.

ΔV1 = I * R1

= (0.72 A) * (2.0 Ω)

= 1.44 V

ΔV2 = I * R2

= (0.72 A) * (4.0 Ω)

= 2.88 V

ΔV3 = I * R3

= (0.72 A) * (6.0 Ω)

= 4.32 V

Answer:

• The potential difference across resistor R1 is 1.44 V
• The potential difference across resistor R2 is 2.88 V
• The potential difference across resistor R3 is 4.32 V

(d) The power dissipated by each resistor can be calculated using the equation P = I^2 * R, where I is the current flowing through the resistor and R is the resistance of the resistor.

P1 = I^2 * R1

= (0.72 A)^2 * (2.0 Ω)

= 0.5184 W

P2 = I^2 * R2

= (0.72 A)^2 * (4.0 Ω)

= 2.0736 W

P3 = I^2 * R3

= (0.72 A)^2 * (6.0 Ω)

= 4.4288 W

Answer:

The power dissipated by resistor R1 is approximately 0.5184 W.

The power dissipated by resistor R2 is approximately 2.0736 W.

The power dissipated by resistor R3 is approximately 4.4288 W.