Answer: If one bulb goes out the other bulbs stay lit.
If there is a break in one branch of the circuit, current can still flow through the other branches.
Explanation:
I think it would be yes because the drum is submerged in water and the water would slow the sound waves, making the sound softer. Right?
Answer:
1. 12 V
2a. R₁ = 4 Ω
2b. V₁ = 4 V
3a. A = 1.5 A
3b. R₂ = 4 Ω
4. Diagram is not complete
Explanation:
1. Determination of V
Current (I) = 2 A
Resistor (R) = 6 Ω
Voltage (V) =?
V = IR
V = 2 × 6
V = 12 V
2. We'll begin by calculating the equivalent resistance. This can be obtained as follow:
Voltage (V) = 12 V
Current (I) = 1 A
Equivalent resistance (R) =?
V = IR
12 = 1 × R
R = 12 Ω
a. Determination of R₁
Equivalent resistance (R) = 12 Ω
Resistor 2 (R₂) = 8 Ω
Resistor 1 (R₁) =?
R = R₁ + R₂ (series arrangement)
12 = R₁ + 8
Collect like terms
12 – 8 =
4 = R₁
R₁ = 4 Ω
b. Determination of V₁
Current (I) = 1 A
Resistor 1 (R₁) = 4 Ω
Voltage 1 (V₁) =?
V₁ = IR₁
V₁ = 1 × 4
V₁ = 4 V
3a. Determination of the current.
Since the connections are in series arrangement, the same current will flow through each resistor. Thus, the ammeter reading can be obtained as follow:
Resistor 1 (R₁) = 4 Ω
Voltage 1 (V₁) = 6 V
Current (I) =?
V₁ = IR₁
6 = 4 × I
Divide both side by 4
I = 6 / 4
I = 1.5 A
Thus, the ammeter (A) reading is 1.5 A
b. Determination of R₂
We'll begin by calculating the voltage cross R₂. This can be obtained as follow:
Total voltage (V) = 12 V
Voltage 1 (V₁) = 6 V
Voltage 2 (V₂) =?
V = V₁ + V₂ (series arrangement)
12 = 6 + V₂
Collect like terms
12 – 6 = V₂
6 = V₂
V₂ = 6 V
Finally, we shall determine R₂. This can be obtained as follow:
Voltage 2 (V₂) = 6 V
Current (I) = 1.5 A
Resistor 2 (R₂) =?
V₂ = IR₂
6 = 1.5 × R₂
Divide both side by 1.5
R₂ = 6 / 1.5
R₂ = 4 Ω
4. The diagram is not complete
Answer:
Explanation:
Firstly, when you measure the voltage across the battery, you get the emf,
E = 13.0 V
In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.
Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (
). But this is not the case.
Let the resistance of the ammeter be r
Hence, using Ohm's law we get the following 2 equations:
.......(1)
......(2)
Substituting the value of r from (2) in (1), we have,
which simplifying gives us, 
(which is our required solution)
putting the value of z in either (1) or (2) gives us, r = 0.5325 
Answer:
First of all, “moist air” is air with a high water vapor content. Water vapor, the invisible, gaseous form of water, occurs in highly variable amounts in the atmosphere. Water is composed of a hydrogen atom and two oxygen atoms (H2O) and has a molecular weight of 18 grams per mole.
