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When one bulb burns open (meaning it stops conducting electricity), it effectively becomes an open circuit in the parallel arrangement. In a parallel circuit, each branch operates independently of the others.

Therefore, if one bulb burns open, the other two bulbs will still light up because the open bulb does not affect the operation of the other branches in the parallel circuit.

So, the correct option is: the other two bulbs will light.

To find the current in each branch of a parallel circuit, you can use the formula:

\[ I_{\text{branch}} = \dfrac{V}{R_{\text{branch}}} \]

Where:
- \( I_{\text{branch}} \) is the current in the branch,
- \( V \) is the total voltage across the parallel branches, and
- \( R_{\text{branch}} \) is the resistance of the branch.

Given:
- Total current (\( I_{\text{total}} \)) = 16 A
- Resistance of the first branch (\( R_1 \)) = 8 ohms
- Resistance of the second branch (\( R_2 \)) = 12 ohms

The equivalent resistance (\( R_{\text{eq}} \)) for parallel resistances is given by:

\[ \dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \]

\[ \dfrac{1}{R_{\text{eq}}} = \dfrac{1}{8} + \dfrac{1}{12} \]

\[ \dfrac{1}{R_{\text{eq}}} = \dfrac{3}{24} + \dfrac{2}{24} \]

\[ \dfrac{1}{R_{\text{eq}}} = \dfrac{5}{24} \]

\[ R_{\text{eq}} = \dfrac{24}{5} \]

Now, we can find the total voltage (\( V \)) using Ohm's Law:

\[ V = I_{\text{total}} \times R_{\text{eq}} \]

\[ V = 16 \times \dfrac{24}{5} \]

\[ V = \dfrac{384}{5} \]

Now, we can find the current in each branch:

1. For the first branch:

\[ I_{\text{branch1}} = \dfrac{V}{R_1} = \dfrac{\dfrac{384}{5}}{8} = \dfrac{48}{5} \]

2. For the second branch:

\[ I_{\text{branch2}} = \dfrac{V}{R_2} = \dfrac{\dfrac{384}{5}}{12} = \dfrac{32}{5} \]

So, the current in each branch is approximately:\[ I_{\text{branch1}} \approx 9.6 \, \text{A} \]\[ I_{\text{branch2}} \approx 6.4 \, \text{A} \]

Therefore, the correct option is: 9.6 A, 6.4 A.

The property of a conductor due to which it passes current is called conductance. Conductance is the reciprocal of resistance and is a measure of how easily a material conducts electricity. The unit of conductance is the siemens (S).

The SI unit of power is the watt (W). One watt is defined as one joule per second. It is named after James Watt, the Scottish engineer who contributed to the development of the steam engine.

In a series circuit, the same current flows through each component. Therefore, in the given scenario with four bulbs of 40 W each connected in series, the statement that is true is:

The current through each bulb is the same.

However, it's important to note that the voltage across each bulb is not necessarily the same in a series circuit. The voltage is distributed among the components based on their resistance. Additionally, the power dissipation in each bulb is the same in this case because they are connected in series, and the current through each bulb is the same. Therefore, the correct statement is:

The current through each bulb is the same.

The resistance of a tungsten lamp increases as the applied voltage increases.

This phenomenon is commonly observed in incandescent lamps, like those made of tungsten. As the voltage across the lamp increases, the temperature of the filament increases, causing the resistance of the filament to increase as well. This is known as positive temperature coefficient of resistance.

Therefore, the correct statement is:

Increases

The resistance of a conductor is directly proportional to its length and inversely proportional to its cross-sectional area. 

Let's denote the resistance of the conductor with the smaller cross-sectional area as \( R_1 \) and the resistance of the conductor with the larger cross-sectional area as \( R_2 \). We're given that the resistance of \( R_1 \) is 40 ohms and the cross-sectional area of \( R_2 \) is four times that of \( R_1 \).

Since the length of both conductors is the same, the resistance is directly proportional to the cross-sectional area. Therefore, if the cross-sectional area of \( R_2 \) is four times that of \( R_1 \), the resistance of \( R_2 \) will be one-fourth of \( R_1 \).

\[ R_2 = \dfrac{1}{4} \times 40 = 10 \, \text{ohms} \]

So, the resistance of the conductor with the larger cross-sectional area will be 10 ohms.

When a parallel circuit is opened in the main line (meaning the main line is interrupted), the circuit is effectively broken, and no current flows through any branch. This is because in a parallel circuit, each branch provides a separate path for current to flow. If the main line is opened, there is no complete path for current to flow through any branch.

Therefore, the correct statement is:

The current is zero in all branches.

The statement that is true for both a series and a parallel DC circuit is:

Currents are additive.

In a series circuit, the same current flows through all elements, and in a parallel circuit, the total current is the sum of the individual branch currents. Therefore, the principle of current being additive holds true for both series and parallel circuits.

To find the voltage across a resistor, you can use Ohm's Law, which states that voltage (\(V\)) is equal to current (\(I\)) multiplied by resistance (\(R\)):

\[ V = I \times R \]

Given:
Resistance (\(R\)) = 33 ohms
Current (\(I\)) = 2 A

Substituting the given values into the formula:

\[ V = 2 \, \text{A} \times 33 \, \Omega \]\[ V = 66 \, \text{V} \]

Therefore, the voltage across the resistor is 66 V.