If Resistors are in series= The equivalent is the sum.
E.g R1 and R2 in series, R = R1 + R2.
If in Parallel, equivalent is Product/sum.
E.g If R1 and R2 in parallel, R = (R1*R2)/(R1+R2)
1.) 60 is parallel with 40 and both are then in series with 20.
60//40 = (60*40)/(60+40) = 2400/100 = 24
Now the 24 is in series with the 20
R = 24 + 20 = 44 ohms.
2.) 80 is in series with 40 and both are then in parallel with 40.
Solving the series, R = 80 + 40 =120.
Parallel: 120//40 = (120*40)/(120+40) = 4800/160 = 30
Equivalent Resistance = 30 ohms.
3.) 100 is in parallel with 100 and both are then in series with the parallel of 50 and 50.
The 1st parallel = (100*100)/(100+100) = 10000/200 = 50
The 2nd parallel = (50*50)/(50+50) = 2500/100 = 25.
Solving the series = 50 + 25 =75 ohms.
Cheers.
Through Shannon's Theorem, we can calculate the capacity of the communications channel using the value of its bandwidth and signal-to-noise ratio. The capacity, C, can be expressed as
C = B × log₂(1 + S/N)
where B is the bandwidth of the channel and S/N is its signal-to-noise ratio.
Since the given SN ratio is in decibels, we must first express it as a ratio with no units as
SN (in decibels) = 10 × log (S/N)
30 = 10log(S/N)
log(S/N) = 3
S/N = 10³ = 1000
Now that we have S/N, we can solve for its capacity (in bits per second) as
C = 4000 × log₂(1 + 1000)
C = 39868.91 bps
Thus, the maximum capacity of the channel is 39868 bps or 40 kbps.
Answer: 40 kbps
Answer:
- 0.3sin6000t A
Explanation:
Voltage, v = 10 cos 6000t V
Capacitance = 5-uF
Current flowing through, i(t)
i(t) = c * d/dt (V)
c = 5-uF = 5 * 10^-6 F
i(t) = (5 * 10^-6) * d/dt(10 cos 6000t)
d/dt(10 cos 6000t) = (10 * 6000) * (-sin 6000t)
Hence,
i(t) = (5*10^-6) * (10*6000) * (-sin 6000t)
i(t) = 5*10^-6 * 6*10^4 * - sin6000t
i(t) = 30 * 10^-2 * - sin6000t
i(t) = 0.3*-sin6000t
i(t) = - 0.3sin6000t Ampere
Answer:
<h3>13,976.23Joules</h3>
Explanation:
Workdone by the rope is expressed using the formula;
W = Fd sin(theta)
F is the tension in the rope = 180
d is the displacement = 300m
theta is the angle of inclination = 15°
Substitute the given parameters into the formula;
W = 180(300)sin15
W = 54000sin 15
W = 13,976.23
Hence the workdone by the rope is 13,976.23Joules
