An 8 Ω resistor is connected to a 6 V battery and another resistor in series. If the current in the circuit is 0.5 A, what is th
1 answer:
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Answer:
Decreases, Increases
Explanation:
Resistance is parallel can be calculated using
1/Req = 1/R1 + 1/R2 + 1/R3 +....
Then, as more resistor is added in parallel the equivalent resistance is reduced.
Let use a simple sample
Let all the resistor have equal resistances
Let say R = R1 = R2 = R3 =...Rn
Then, 1/Req = 1/R1 + 1/R2 + 1/R3 +....
1/Req = 1/R + 1/R + 1/R +.... 1/Rn
Req = R/n
Check attachment on how I got that.
This implies that, the equivalent resistance will always be less than the original resistance, since n>1
So, as n increases (I.e. as the number of resistance increases), the equivalent resistance reduces.
B. Now, to know if the current reduces or increases
Using Ohms law
V = iR
Then, I = V/R
So, let assume the voltage is constant, then, the current is inversely proportional to the resistance, so as we know that the resistance is reducing, then the current will be increasing.
So current increase as we add more resistor in parallel to a circuit
Answer:
After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec
Explanation:
Area of a circle = πr²
where;
r is the circle radius
Differentiate the area with respect to time.
dr/dt = 4 ft/sec
after 12 seconds, the radius becomes =
To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt
dA/dt = 1206.528 ft²/sec
Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec