# I want to know the equivalent resistance , the flow , the tension and the developing power

Equivalent resistance:

Resistors R2 ad R3 are connected in parallel to each other.

Equivalent resistance of R2 and R3 is given as:

1/R23 = 1/R2 + 1/R3 = (R2+R3)/R2R3

or, R23 = (R2R3)/(R2+R3)

and R23 is in series with R1.

Therefore, equivalent resistance of the circuit is given as:

**R123 = R1 + R23 = R1 + (R2R3)/(R2+R3)**

**Current:**

The current through R1 and R23 (equivalent resistance of R2 and R3) will be the same, as `I_1` . The current will split between R2 and R3 (values are given as I2 and I3, respectively).

further, `I_1` = `I_2 + I_3`

**Voltage:**

The same voltage will pass through R2 and R3. Also, the voltage of the battery will be split between R1 and R23 (equivalent resistance of R2 and R3).

**Power:**

Power of the circuit can be calculated as I^2R, where I is the current flowing through the circuit and R is the equivalent resistance of the circuit (calculated earlier).

Therefore, Power = `I_1^2 (R_1 + (R_2R_3)/(R_2+R_3))`

Hope this helps.

In the circuit given

The equivalent resistance is calculated as follows

R equal= R1+ parallel resistors (R2,R3)

= R1 + `(R2*R3)/(R2+R3)`

As when the resistors are conneted in parallel then the equivalent resistance is equal to `1/(R2)+ 1/(R3)= (R2*R3)/(R2+R3)`

and now in the circuit, let the total current be I total

as the two resistors are connected in parallel the current is divided into I1 and I2 which is shown in the circuit

so the `ITotal` = I2 +I3

So, now we can directly calculate the power (P) by the equation

P = (Itotal)^2 * (Requal)

= `(I2+I3)^2 * (R1 + (R2*R3)/(R2+R3))`

is the total power developed