In this parallel circuit, there are two branches:

• Branch 1: Two bulbs, each with 10 Ω resistance, connected in series.
• Branch 2: One bulb with 10 Ω resistance.

Step 1: Calculate the resistance of Branch 1

Since the two bulbs in Branch 1 are connected in series, their total resistance $R_1$ is: $R_1 = R + R = 10 \, \Omega + 10 \, \Omega = 20 \, \Omega$

Step 2: Resistance of Branch 2

Branch 2 has a single bulb with 10 Ω resistance: $R_2 = 10 \, \Omega$

Step 3: Calculate the total resistance of the parallel circuit

The total resistance $R_{\text{total}}$ in a parallel circuit is given by: $\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2}$ Substitute the values: $\frac{1}{R_{\text{total}}} = \frac{1}{20} + \frac{1}{10}$ Find the common denominator and sum: $\frac{1}{R_{\text{total}}} = \frac{1}{20} + \frac{2}{20} = \frac{3}{20}$ Now, solve for $R_{\text{total}}$: $R_{\text{total}} = \frac{20}{3} \, \Omega \approx 6.67 \, \Omega$

Final Answer:

The total resistance of the circuit is approximately $6.67 \, \Omega$.

Would you like more details or have any questions?

Here are 5 related questions for practice:

1. What is the voltage across the whole circuit if the total current is 4.5 A?
2. How much current flows through each branch of the parallel circuit?
3. If another bulb is added to Branch 2, how will the total resistance change?
4. How do you calculate the total resistance of more than two branches in a parallel circuit?
5. What happens to the total current if the resistance in Branch 1 increases?

Tip: In a parallel circuit, the total resistance is always less than the smallest individual resistance!