Question

1. Two resistors R₁ (12 ohm) and R₁ (24 ohm) are

connected in series across a 6.0 V battery
of negligible internal resistance.
I
Draw a circuit diagram (to the right) and calculate:
The total resistance of the two resistors:
The total current flowing in the circuit:
The current flowing in R₁
The current flowing in R2
The total power consumed by R₁ and R₂
||
S

Answer 1

To calculate the total resistance of the two resistors in series, we add them:

R_total = R₁ + R₂ = 12 ohm + 24 ohm = 36 ohm

To calculate the total current flowing in the circuit, we use Ohm's Law:

I = V / R_total = 6.0 V / 36 ohm = 0.17 A

The current flowing in R₁ is the same as the total current, since they are in series:

I₁ = 0.17 A

The current flowing in R₂ is also the same, since they are in series:

I₂ = 0.17 A

To calculate the total power consumed by R₁ and R₂, we use the formula:

P = I²R

For R₁:

P₁ = I₁²R₁ = (0.17 A)²(12 ohm) = 0.35 W

For R₂:

P₂ = I₂²R₂ = (0.17 A)²(24 ohm) = 0.71 W

The total power consumed by both resistors is:

P_total = P₁ + P₂ = 0.35 W + 0.71 W = 1.06 W

Answer 2

1. The total resistance of the two resistors:

Since the resistors are connected in parallel.

Total resistance = R₁ + R2

Total resistance = 12 ohm + 24 ohm

Total resistance = 36 ohms

2. The total current flowing in the circuit:

I = V/R

where ,

• I is current,
• V is voltage (6.0 V)
• and r is resistance.

I = 6.0 V /36 ohm

I = 0.17 amperes

3. The current flowing in R₁

Since he two resistance R₁ and R₂ are connected in series. The current flowing through these two will be equal to the total current flowing in the circuit.

Current flowing in R₁ = 0.17 A

4. The current flowing in R2 = total current flowing in the circuit:

The current flowing in R₂ = 0.17 A.

5. The total power consumed by R₁ and R₂

power consumed by R₁

P = I²R

P = (0.17)²/12

P = 0.0289/12

P = 0.35 (approx)

Now, power consumed by R₂

P = (0.17)² × 24

P = 0.0289 × 24

P = 0.70 (approx)

Total power = power consumed by R₁ + R₂

Total power = 0.35 + 0.70

Total power consumed = 1.05 W