A visually engaging and educational image that represents the concept of measuring electrical values. The image has a detailed representation of a coil connected to an electrical supply and a meter measuring values. The coil is intricately drawn with a metallic sheen and closely packed wire turns. An electrical current - one being direct and another alternating - of different values is illustrated visually with distinct sine waves. The frequency, current, and potential difference values are indicated symbolically. The image is vibrant but does not contain any text.

The potential difference measured across a coil is 20 V

when a direct current of 2 A is passed through it. With
an alternating current of 2 A at 40 Hz, the p.d. across
the coil is 140 V. If the coil is connected to a 230 V,
50 Hz supply, calculate: (a) the current; (b) the active
power; (c) the power factor

R = E/I = 20/2 = 10 Ohms.

Xl = E/I = 140/2 = 70 Ohms @ 40 Hz.= The
inductive reactance.

Xl == 2pi*F*L = 70 Ohms.
6.28*40*L = 70.
L = 0.279 Henrys. = The inductance.

a. Xl = 2pi*F*L = 6.28*50*0.279 = 87.5
Ohms.
I = E/Xl = 230/87.5 = 2.63 Amps.

b. Tan A = Xl/R = 87.5/10 = 8.75.
A = 83.5o = Phase shift angle.

P = E*I*Cos A =

Why did the coil bring a calculator to the party? Because it wanted to solve these electrifying problems!

(a) To find the current, we can divide the p.d. (voltage) by the impedance of the coil. Given that the p.d. is 140 V and the current is 2 A, we can calculate the impedance using Ohm's Law, which states that impedance equals voltage divided by current. So, 140 V / 2 A gives us an impedance of 70 ohms.

Now, let's calculate the current when the coil is connected to a 230 V, 50 Hz supply. Similar to before, we can use Ohm's Law. The impedance of the coil remains the same, so the current will be 230 V / 70 ohms, which is approximately 3.29 A.

(b) The active power can be calculated using the formula P = IV, where P is the power, I is the current, and V is the voltage. For the given supply voltage of 230 V and the current of 3.29 A, the active power will be 230 V * 3.29 A, which is approximately 757.7 W.

(c) To calculate the power factor, we need to divide the active power by the apparent power. The apparent power can be found using the formula S = IV, where S is the apparent power. For the given supply voltage of 230 V and the current of 3.29 A, the apparent power will be 230 V * 3.29 A, which is approximately 757.7 VA (volt-amperes).

Therefore, the power factor is 757.7 W / 757.7 VA, which means the power factor is 1.

Hope that sparks a bit of humor in your calculations!

To solve this problem, we'll use the formulas for impedance, power, and power factor in an AC circuit.

(a) First, we need to find the impedance of the coil. We can use the formula:

Impedance (Z) = Voltage (V) / Current (I)

For the direct current:
Z_dc = 20 V / 2 A = 10 Ω

For the alternating current:
Z_ac = 140 V / 2 A = 70 Ω

Now, let's calculate the current in the coil when connected to a 230 V, 50 Hz supply.

Impedance (Z) = Voltage (V) / Current (I)
230 V = Z × I
I = 230 V / Z

For a 50 Hz supply:
I = 230 V / 70 Ω = 3.29 A

Therefore, the current flowing through the coil is 3.29 A.

(b) To calculate the active power (P), we'll use the formula:

Active Power (P) = Voltage (V) × Current (I) × Power Factor (PF)

Since the power factor is not given, we'll need to determine it. We can find the power factor by using the formula:

Power Factor (PF) = Active Power (P) / Apparent Power (S)

We already have the active power, so we need to find the apparent power (S) first.

Apparent Power (S) = Voltage (V) × Current (I)

For the 50 Hz supply:
S = 230 V × 3.29 A = 756.7 VA

Now we can calculate the power factor:

PF = P / S = P / (V × I) = P / (230 V × 3.29 A)

Assuming the power factor is cos(θ), we can rewrite the equation as:

cos(θ) = P / (230 V × 3.29 A)

Plug in the values to find the power factor:

cos(θ) = P / (230 V × 3.29 A)

(c) Now, let's calculate the power factor.

Power factor = cos(θ) = P / (230 V × 3.29 A) = (P / (230 V × 3.29 A)) × 100 (to convert it to a percentage)

Assuming you have the value for active power (P), plug it into the formula to find the power factor.

To calculate the current, active power, and power factor of the coil connected to a 230 V, 50 Hz supply, we need to use the information given. Let's break down each part of the question and solve them one by one.

(a) Calculate the Current:
We know that the potential difference across the coil is 140 V when an alternating current of 2 A at 40 Hz is passed through it. However, we need to find the current when the coil is connected to a 230 V, 50 Hz supply.

To find the current, we can use the formula for the impedance of an inductive load. Inductive impedance (Z_L) is given by:
Z_L = V_L / I_L

Where:
Z_L is the impedance of the coil,
V_L is the potential difference across the coil, and
I_L is the current through the coil.

Given that the potential difference across the coil is 140 V and the current through the coil is 2 A, we can substitute these values into the formula:
Z_L = 140 V / 2 A
Z_L = 70 Ω

Now, when the coil is connected to a 230 V supply, we can calculate the current (I) using Ohm's Law:
I = V / Z

Where:
I is the current,
V is the potential difference, and
Z is the impedance.

Substituting the values:
I = 230 V / 70 Ω ≈ 3.29 A

Therefore, the current through the coil is approximately 3.29 A.

(b) Calculate the Active Power:
To determine the active power, we need to calculate the power consumed by the coil connected to the 230 V supply. Active power (P) is given by:
P = V * I * cos(θ)

Where:
P is the active power,
V is the potential difference,
I is the current, and
θ is the phase angle between the voltage and current.

Since we are dealing with a purely inductive load, the power factor is lagging, and the phase angle is 90 degrees. Therefore, the cos(θ) is 0.

P = 230 V * 3.29 A * 0
P = 0

The active power consumed by the coil is 0 since it is an inductive load.

(c) Calculate the Power Factor:
The power factor (PF) is the cosine of the phase angle between the voltage and current. In this case, since the load is purely inductive, the power factor is 0.

Therefore, the power factor of the coil connected to the 230 V supply is 0.

To summarize:
(a) The current through the coil is approximately 3.29 A.
(b) The active power consumed by the coil is 0.
(c) The power factor of the coil is 0.

A coil takes a current of 10.0 A and dissipates 1410 W

when connected to a 230 V, 50 Hz sinusoidal supply.
When another coil is connected in parallel with it,
the total current taken from the supply is 20.0 A at a
power factor of 0.866. Determine the current and the
overall power factor when the coils are connected in
series across the same supply