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.
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
11

Answer 1

Here's the circuit diagram:

```

+---R₁---R₂---+

| |

- -

(6V) ( )

|

|

|

|

|

|

|

( )

-

```

Calculations:

1. The total resistance of the two resistors:

R_total = R₁ + R₂

= 12 ohm + 24 ohm

= 36 ohm

Answer: 36 ohm

2. The total current flowing in the circuit:

I = V / R_total

= 6.0 V / 36 ohm

= 0.167 A

Answer: 0.167 A

3. The current flowing in R₁:

I₁ = V / R₁

= 6.0 V / 12 ohm

= 0.5 A

Answer: 0.5 A

4. The current flowing in R₂:

I₂ = V / R₂

= 6.0 V / 24 ohm

= 0.25 A

Answer: 0.25 A

5. The total power consumed by R₁ and R₂:

P = I² * R_total

= (I₁ + I₂)² * R_total

= (0.5 A + 0.25 A)² * 36 ohm

= 0.1875 * 36 ohm

= 6.75 W

Answer: 6.75 W