Applying Kirchhoff’s Junction Rule, what happens to the power source and current when the parallel circuit has two branches, eac
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Hi there! :)
Reference the diagram below for clarification.
1.
We must begin by knowing the following rules for resistors in series and parallel.
In series:
In parallel:
We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.
Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
2.
We can use Ohm's law to solve for the current in the circuit.
3.
For resistors in series, both resistors receive the SAME current.
Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.
In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.
4.
Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.
Answer:
a
The orbital speed is
b
The escape velocity of the rocket is
Explanation:
Generally angular velocity is mathematically represented as
Where T is the period which is given as 1.6 days =
Substituting the value
At the point when the rocket is on a circular orbit
The gravitational force = centripetal force and this can be mathematically represented as
Where G is the universal gravitational constant with a value
M is the mass of the earth with a constant value of
r is the distance between earth and circular orbit where the rocke is found
Making r the subject
The orbital speed is represented mathematically as
Substituting value
The escape velocity is mathematically represented as
Substituting values