The circuit to the right consists of a battery ( V 0 = 64.5 V) (V0=64.5 V) and five resistors ( R 1 = 711 (R1=711 Ω, R 2 = 182 R2=182 Ω, R 3 = 663 R3=663 Ω, R 4 = 534 R4=534 Ω, and R 5 = 265 R5=265 Ω). Find the current passing through each of the specified points. -g

The current in R₁ is 0.0816 A.

The current at H point is 0.0243 A.

Explanation:

Given that,

Voltage = 64.5

Resistance is

Suppose, The specified points are R₁ and H.

According to figure,

R₂,R₃,R₄ and R₅ are connected in parallel

We need to calculate the resistance

Using parallel formula

Put the value into the formula

R and R₁ are connected in series

We need to calculate the equilibrium resistance

Using series formula

We need to calculate the equivalent current

Using ohm's law

Put the value into the formula

We know that,

In series combination current distribution in each resistor will be same.

So, Current in R and R₁ will be equal to .

The current at h point will be equal to current in R₅

We need to calculate the voltage in R

Using ohm's law

Put the value into the formula

In resistors parallel combination voltage distribution in each part will be same.

So,

We need to calculate the current at H point  

Using ohm's law

Put the value into the formula

Hence, The current in R₁ is 0.0816 A.

The current at H point is 0.0243 A.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .