Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
Answer:
Explanation:
We can find the resistance of the wire by using Ohm's law:
where
V is the voltage applied
R is the resistance
I is the current
In this problem, we know I = 6 A and V = 68 V, so we can re-arrange the equation to find the resistance of the wire:
Answer:
λ = 596 nm.
Explanation:
Fringe width = λ D / d
λ is wave length , D is screen distance and d is slit separation.
Putting the values
1.62 x 10⁻² =( λ x 5.3 ) / .195 x 10⁻³
λ = 596 nm.
Answer:
The equation which describes conservation of charge is 
Explanation:
The law of conservation charge states that for an isolated system that sum of initial charges is equal to sum of final charges, that is the total charge is conserved.
let the sum of initial charges = 
let the sum of the final charges = 
Therefore, the equation which describes conservation of charge is
