The current passing through a circuit consisting of a battery of 12 V and resistor of 2 ohms is 6 Ampere .
Explanation:
- Assume the wires are ideal with zero resistance.
- The current passing through the circuit will be
I = V/R = 12/2 = 6.000 A.
In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li
1 answer:
Answer:
Total energy saving will be 0.8 KWH
Explanation:
We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW
30 bulbs are of power 60 W
So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW
Total power of 80 bulbs = 1.8+5 = 6.8 KW
Total time = 3 hour
We know that energy
Now power of each CFL bulb = 25 W
So power of 80 bulbs = 80×25 = 2000 W = 2 KW
Energy of 80 bulbs = 2×3 = 6 KWH
So total energy saving = 6.8-6 = 0.8 KWH
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The electric field at one corner of a square is 1614217 N/C.
Explanation:
The distance between x and y direction diagonals.
As per the given details the distance between diagonals is calculated as
0.5² + 0.5² = c² => c = 0.707 m
Charge to the right: In x direction
In order to find the electric charge towards x direction
we use e = kq/r² formula
As 'k' is coulomb's constant it's value is 9 x N m²/C²
e = (9 x )(250 x
) / (0.5)²
e = 9 x N/C
Charge diagonal:
e = kq/r²
e = [(9 x )(250 x
) / (0.707)²] cos 45
e = 225000√2 N/C
X direction sum = 1218198 N/C.
Similarly as shown in x direction the charge is same for y direction also
Charge below: For y direction
e = kq/r²
e = (9 x )(250 x
) / (0.5)²
e = 9 x N/C
Charge diagonal:
e = kq/r²
e = [(9 x )(250 x
) / (0.5)²] sin 45
e = 159099 N/C
Y direction sum = 1059099 N/C
Resultant electric field strength:
1218198 ² + 1059099² = e²
e = 1614217 N/C [45 degrees below the horizontal]