Let's solve each part of this circuit problem step-by-step.

Step 1: Calculate Total Resistance

Since the light bulbs are in parallel, we use the parallel resistance formula:

$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

where:

• $R_1 = 30 \Omega$ (top),
• $R_2 = 24 \Omega$ (middle),
• $R_3 = 12 \Omega$ (bottom).

Calculating:

$\frac{1}{R_{\text{total}}} = \frac{1}{30} + \frac{1}{24} + \frac{1}{12}$

$\frac{1}{R_{\text{total}}} = 0.0333 + 0.0417 + 0.0833 = 0.1583$

$R_{\text{total}} = \frac{1}{0.1583} \approx 6.32 \Omega$

So, the total resistance $R_{\text{total}} \approx 6.32 \Omega$.

Step 2: Calculate Current in Each Branch with a 12V Battery

The current in each branch can be found using Ohm’s law $I = \frac{V}{R}$.

1. Bottom Branch (12 Ω): $I_{\text{bottom}} = \frac{12\text{ V}}{12\Omega} = 1\text{ A}$

2. Middle Branch (24 Ω): $I_{\text{middle}} = \frac{12\text{ V}}{24\Omega} = 0.5\text{ A}$

3. Top Branch (30 Ω): $I_{\text{top}} = \frac{12\text{ V}}{30\Omega} = 0.4\text{ A}$

Summary of Answers

1. Total Resistance: $R_{\text{total}} \approx 6.32 \Omega$
2. Current on Bottom Branch: $I_{\text{bottom}} = 1 \text{ A}$
3. Current on Middle Branch: $I_{\text{middle}} = 0.5 \text{ A}$
4. Current on Top Branch: $I_{\text{top}} = 0.4 \text{ A}$

Would you like more details on any step?

Related Questions

1. What would be the total current supplied by the battery?
2. How would the total resistance change if an additional 15Ω bulb were added in parallel?
3. What would happen to the currents if the battery voltage were increased to 24V?
4. How can we calculate the power dissipated in each branch?
5. What if the resistors were in series instead of parallel; how would that affect the total resistance?

Tip

For circuits in parallel, remember that adding more resistors always decreases the total resistance.