What problem do refractor telescopes have that reflectors don't? Group of answer choices bad seeing light loss from secondary el
1 answer:
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<h2>Answer:</h2>
(a) Attached to the response as Figure 1.
(b) 35.0Ω
(c) Across 5.0Ω = 1.3V
Across 10.0Ω = 2.6Ω
Across 20.0Ω = 5.2Ω
<h2>Explanation:</h2>(a) The labelled circuit using the correct symbols (for the resistors and battery) has been attached to this response.
(b) Since the resistors are hooked up in series, their equivalent resistance R, is found by adding the individual resistances of the resistors (R₁, R₂ and R₃). i.e
R = R₁ + R₂ + R₃ -------------------(i)
Where;
R₁ = 5.0 Ω
R₂ = 10.0 Ω
R₃ = 20.0 Ω
<em>Substitute these values into equation (i) as follows;</em>
∴ R = 5.0 Ω + 10.0 Ω + 20.0 Ω
∴ R = 35.0 Ω
Therefore, the equivalent resistance is ∴ R = 35.0Ω
(c) When resistors are connected in series, the same current passes through them. To get the current through each resistor;
i. First, replace the resistors by their equivalent resistor as calculated above. The diagram has been attached to this response.
ii. As seen in the diagram, the current flowing through the equivalent resistor can be calculated using Ohm's law as follows;
V = I R ------------------(ii)
Where;
V = Voltage supplied to the circuit = 9.0V
I = Current through the circuit
R = Resistance of the equivalent resistor = 35.0Ω
Substitute these values into equation (ii)
9.0 = I x 35.0
I =
I = 0.26A
This is also the current flowing through each of the resistors separately.
iii. Calculate the voltage drop across
1.<em> 5.0 Ω resistor</em>
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 5.0Ω resistor
I = current through the 5.0Ω resistor = 0.26A
R = resistance of the 5.0Ω resistor = 5.0Ω
=> V = 0.26 x 5.0
=> V = 1.3V
2.<em> 10.0 Ω resistor</em>
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 10.0Ω resistor
I = current through the 10.0Ω resistor = 0.26A
R = resistance of the 10.0Ω resistor = 10.0Ω
=> V = 0.26 x 10.0
=> V = 2.6V
3.<em> 20.0 Ω resistor</em>
Applying Ohm's law from equation (ii)
V = I x R
Where;
V = voltage drop across the 20.0Ω resistor
I = current through the 20.0Ω resistor = 0.26A
R = resistance of the 20.0Ω resistor = 10.0Ω
=> V = 0.26 x 20.0
=> V = 5.2V