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:
Potential energy.
Explanation:
Batteries are a form of chemical energy which is stored in the bonds of molecules contained within the battery acid, batteries are a form of potential energy making chemical potential energy too.
The answer is C. Hope it helps
Answer:
Explanation:
Given
Initial Intensity of light is S
when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.
When it is passed through a second Polarizer with its transmission axis 
here 
When it is passed through third Polarizer with its axis 
to first but 
to second thus 
When middle sheet is absent then Final Intensity will be zero
The primary coil has more turns than the secondary coil. so D.
