Answer:
Explanation:
side of the square loop, a = 7 cm
distance of the nearest side from long wire, r = 2 cm = 0.02 m
di/dt = 9 A/s
Integrate on both the sides
i = 9t
(a) The magnetic field due to the current carrying wire at a distance r is given by
(b)
Magnetic flux,
(c)
R = 3 ohm
magnitude of voltage is
e = 1.89 x 10^-7 V
induced current, i = e / R = (1.89 x 10^-7) / 3
i = 6.3 x 10^-8 A
Answer:
a-
V= IR
9V = I ×( 12+6)
I = 9/ 18 A = 0.5 A
b
V=IR
240 = 6 A ×( 20 + R)
40 = 20 + R
R = 20 ohm
c
resultant resistance of the 2 parallel resistances= Ro
1/Ro = 1/ 5 + 1/ 20
1/Ro =( 20+5)/100
= 1/Ro = 1/4
Ro= 4 ohm
V=IR
V = 2A × ( 1+ 4 OHM)
V = 10V
d
equivalent resistance = Ro
1/Ro = 1/(2+8) + 1/(5+5)
1/Ro = 1/10 +1/10
2/10 = 1/ Ro
Ro= 10/2 = 5 ohm
V = IR
12V = I × 5Ohm
I=2.4 A
B. Energy
A power company charges its customers for electricity based upon B. Energy.
<h3>
Explanation:</h3>
Kilo-watt Hours (kWh) is the unit that measures the electricity consumption of customers. Since Power is defined as the rate at which electrical energy is transferred by an electrical circuit per unit time,
If energy is transmitted at a constant rate over a period of time, the total energy in kilowatt hours is the product of power in kilowatts(kW) and time in hours (h)