when a substance such as ice melts, its temperature increases. desrcibe what happens to the arrangment of water molecules as the
1 answer:
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<h2>Answer:</h2>
0.46Ω
<h2>Explanation:</h2>The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;
E = V + Ir --------------------(a)
Where;
I = current flowing through the circuit
But;
V = I x Rₓ ---------------------(b)
Where;
Rₓ = effective or total resistance in the circuit.
<em>First, let's calculate the effective resistance in the circuit:</em>
The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.
Let;
R₁ = resistance in the first bulb
R₂ = resistance in the second bulb
Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;
P =
=> R = -------------------(ii)
Where;
P = Power of the bulb
V = voltage across the bulb
R = resistance of the bulb
To get R₁, equation (ii) can be written as;
R₁ = --------------------------------(iii)
Where;
V = 12.0V
P = 4.0W
Substitute these values into equation (iii) as follows;
R₁ =
R₁ =
R₁ = 36Ω
Following the same approach, to get R₂, equation (ii) can be written as;
R₂ = --------------------------------(iv)
Where;
V = 12.0V
P = 4.0W
Substitute these values into equation (iv) as follows;
R₂ =
R₂ =
R₂ = 36Ω
Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;
=
+
-----------------(v)
Substitute the values of R₁ and R₂ into equation (v) as follows;
=
+
=
Rₓ =
Rₓ = 18Ω
The effective resistance (Rₓ) is therefore, 18Ω
<em>Now calculate the current I, flowing in the circuit:</em>
Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;
11.7 = I x 18
I =
I = 0.65A
<em>Now calculate the battery's internal resistance:</em>
Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;
12.0 = 11.7 + 0.65r
0.65r = 12.0 - 11.7
0.65r = 0.3
r =
r = 0.46Ω
Therefore, the internal resistance of the battery is 0.46Ω
Answer:
205 V
V = 2.05 V
Explanation:
L = Inductance in Henries, (H) = 0.500 H
resistor is of 93 Ω so R = 93 Ω
The voltage across the inductor is
w = 500 rad/s
IwL = 11.0 V
Current:
I = 11.0 V / wL
= 11.0 V / 500 rad/s (0.500 H)
= 11.0 / 250
I = 0.044 A
Now
V = IR
= (0.044 A) (93 Ω)
V = 4.092 V
Deriving formula for voltage across the resistor
The derivative of sin is cos
V = V
cos (wt)
Putting V = 4.092 V and w = 500 rad/s
V = V
cos (wt)
= (4.092 V) (cos(500 rad/s )t)
So the voltage across the resistor at 2.09 x 10-3 s is which means
t = 2.09 x 10⁻³
V = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))
= (4.092 V) (cos (500 rads/s)(0.00209))
= (4.092 V) (cos(1.045))
= (4.092 V)(0.501902)
= 2.053783
V = 2.05 V