Answer: 25 Ohms
Explanation:
From this question, the following parameters are given:
Voltage V = 1.5 v
Current I = 0.03A
From Ohm's law;
V = IR
Where R = resultant resistance of the two resistors.
Substitute V and I into the formula and make resultant R the subject of formula.
1.5 = 0.03 × R
R = 1.5/0.03
R = 50 Ohms
From the question, it is given that Thr two equal resistors are connected in series.
R = R1 + R2
But R1 = R2
50 = 2R1
R1 = 50/2
R1 = 25
R1 = R2 = 25 Ohms
Therefore, the resistors must each have a value of 25 Ohms
The people are using a lot of electricity blow drying to many peoples hair so i would make a schedule so it dosent get to busy with costumers
Answer:
P.E. = -0.449 J
Explanation:
Potential energy of a charge particle in any electrostatic field is defined as the amount of work done ( in negative ) to bring that charge particle from any position to a new position r.
Now Potential energy is defined by this formula,
P.E. = k q₁ q₂/ r
where P.E. is the potential energy.
k = 1/( 4πε₀) = 8.99 × 10⁹ C²/ ( Nm²)
q₁ = charge of one particle = +1.0μC
q₂ = charge of another particle = -5.0μC
r = distance = 0.1 m
Now , P.E. = 8.99 × 10⁹C²/ ( Nm²) * ( -5.0 × 10⁻⁶ C ) × ( 1 × 10⁻⁶ C ) / 0.1 m
P.E. = -0.449 J
Answer:
in the parallel connection the light bulbs shine less than in the series connection
Explanation:
In a series circuit the current through the whole circuit is the same, therefore the power (brightness) of each bulb is
P = i² R
where R is the resistance of each bulb and i the current of the circuit.
If we connect the light bulbs and the cells in parallel, the current in the circuit is the sum of the east that passes through each light bulb,
i = i₁ + i₂
if the two light bulbs are the same
i = 2 i₁
i₁ = i / 2
so the power of each bulb is is
P = i₁² R
P = R i² / 4
P = ¼ P_initial
Therefore we see that in the parallel connection the light bulbs shine less than in the series connection
