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