The I3 will be 158 A.
<h3>How to find the current through the circuit?</h3>
- The foundation of circuit analysis is Kirchhoff's circuit laws.
- We have the fundamental instrument to begin studying circuits with the use of these principles and the equation for each individual component (resistor, capacitor, and inductor).
- These rules aid in calculating the current flow in various network streams as well as the electrical resistance of a complicated network, or impedance in the case of AC.
To calculate I3 firstly, V4 has to be calculated,
For I3,
Hence, the current through I3 will be 158 A.
To learn more about Kirchoff's laws refer to:
brainly.com/question/86531
#SPJ 10
I would use a resistor rated for 1 W or more. Not less.
<span>Answer:
F(x) = ax^2 - bx
or
F(x) = ax² - bx
F(x) = 30x² - 6x
â«F(x)dx = â«(30x² - 6x)dx
as this is evaluated from zero to x
W = 10x³ - 3x² <===ANS
W = 10(0.42³) - 3(0.42²) - [10(0³) - 3(0²)]
W = 0.212 J <===ANS
W = 10(0.72³) - 3(0.72²) - [10(0.42³) - 3(0.42²)]
W = 1.966 J <===ANS</span>
Answer:
a) velocity v = 322.5m/s
b) time t = 19.27s
Explanation:
Note that;
ads = vdv
where
a is acceleration
s is distance
v is velocity
Given;
a = 6 + 0.02s
so,
Remember that
![v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bds%7D%7Bdt%7D%20%5C%5C%5Cfrac%7Bds%7D%7Bv%7D%20%3D%20dt%5C%5C%5Cint%5Climits%5Es_0%20%7B%5Cfrac%7Bds%7D%7B%5Csqrt%7B12s%2B0.02s%5E%7B2%7D%20%7D%20%7D%20%7D%20%5C%2C%20ds%20%3D%20%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt%20%5C%5Ct%3D%20%20%285%5Csqrt%7B2%7D%20%29%20ln%20%20%5Cfrac%7B%7C%20%5Bs%20%2B%20300%20%2B%20%5Csqrt%7B%28s%5E%7B2%7D%20%20%2B%20600s%29%7D%20%5D%20%7C%7D%7B300%7D%20.......2)
substituting s = 2km =2000m, into equation 1
v = 322.5m/s
substituting s = 2000m into equation 2
t = 19.27s
