A coil of inductance 0.04henry and negligible resistance is connected in parallel with a 10ohms resistor, across a 240volts supply of frequency 50hz. Calculate:1) the current through the resistance, 2) the current through the coil, 3)the supply current, 4)the power in the circuit.
I will be happy to critique your thinking.
a) I=V/R=240@0/10=24 amps at 0 phase shift
b) i=V/Z=240@0/Z where z=j2PI*50*.04=4PI@90
i= 240@0/4PI@90=60/PI @ -90
c) supply= a)+b) I would combine them in vector form
d) Power= VdotI=240*IinC *CosThetainC
which will equal 240*24(current in a))
To calculate the values requested, we will need to use the formulas related to inductance, resistance, and power in AC circuits.
1) The current through the resistance (IR):
We can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R):
IR = V / R
IR = 240 V / 10 Ω
IR = 24 A
2) The current through the coil (IL):
The current through the coil can be calculated using the following formula in an AC circuit:
IL = V / XL
XL = 2πfL, where XL is the inductive reactance, f is the frequency in hertz, and L is the inductance in henries.
XL = 2 × π × 50 Hz × 0.04 H
XL ≈ 12.57 Ω
Thus, the current through the coil is:
IL = V / XL
IL = 240 V / 12.57 Ω
IL ≈ 19.12 A
3) The supply current (IS):
In a parallel circuit, the total current is the sum of the individual branch currents:
IS = IR + IL
IS = 24 A + 19.12 A
IS ≈ 43.12 A
4) The power in the circuit (P):
The power in an AC circuit can be calculated using the formula:
P = VI, where V is the voltage and I is the current.
For the resistance:
PR = V × IR
PR = 240 V × 24 A
PR = 5760 W
For the coil:
PL = V × IL
PL = 240 V × 19.12 A
PL = 4596.8 W
Total power in the circuit:
P = PR + PL
P = 5760 W + 4596.8 W
P ≈ 10,356.8 W
Therefore, the results are as follows:
1) The current through the resistance is approximately 24 A.
2) The current through the coil is approximately 19.12 A.
3) The supply current is approximately 43.12 A.
4) The power in the circuit is approximately 10,356.8 W.