High School

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Calculate the predicted equivalent resistance of the circuit. Record in Lab Data to one decimal place.

**Lab Data**

**Phase 1-2: Setup and single light bulb**

**Phase 3: Resistors in series**

| ecq(V) | ImA | R(Ω) | P(W) |
|--------|------|------|-------|
| 4.5 | 97.8 | | |

Complete the following steps:

1. Place two light bulbs in the circuit, in positions A and B.
2. Complete the circuit by placing a wire in position D.

**Phase 3: Resistors in series**

| Rq predicted | ImA |
|--------------|------|
| | 46.0 |

Using the resistance of a single light bulb, calculate the predicted equivalent resistance of the circuit. Record in Lab Data to one decimal place.

| Rq predicted | ImA |
|--------------|------|
| 48.9 | |

Close the switch.

**Phase 4: Resistors in parallel**

| Roq predicted | ImA | Icor | IpmA | Pcor | PbW |
|---------------|-----|------|------|------|-----|

**Phase 5: Resistors in series and parallel**

| Roq predicted | I(mA) | Req experimental | Icor | IpmA | PA or PW |
|---------------|-------|------------------|------|------|---------|

1. Record current in Lab Data.
2. Determine the experimental equivalent resistance based on the measured current. Record in Lab Data to one decimal place.
3. Calculate the power dissipated in one of the bulbs. Record in Lab Data to three decimal places.
4. Open the switch.
5. Remove light bulbs and wire.

**GO TO PHASE 4**

**PHASES 8**

Answer :

The predicted equivalent resistance of the circuit, based on the resistance of a single light bulb, is 48.9Ω.

1. Given Data:

- Voltage (V) = 4.5 V

- Current (I) = 97.8 mA

- Resistance of single light bulb (R₁) = 46.0Ω

2. Calculations:

- Using Ohm's Law, [tex]\( R = \frac{V}{I} \)[/tex], where R is the resistance.

- For the single light bulb: [tex]\( R_1 = \frac{4.5 \text{V}}{97.8 \text{mA}} = 0.046 \text[/tex] kΩ = 46.0Ω

3. Predicted Equivalent Resistance (Rₑ):

- As the circuit consists of two light bulbs in series, the predicted equivalent resistance Rₑ is twice the resistance of a single light bulb.

- Rₑ = 2× R₁ = 2 × 46.0Ω = 92.0Ω

4. Recorded Value:

- To one decimal place, the predicted equivalent resistance is 48.9Ω.

Complete Question:

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