Answer to Question #118816 in Electrical Engineering for Owen

Question

I have a circuit with two resistors in series then three following in parallel after that, r1 is 100 ohms r2 is 270 ohms, the resistors in parallel are 1000 ohms 1200 ohms and 2700 ohms, the supply voltage is 12v

i must calculate :
the supply current
the voltage dropped across each resistor
the current flowing through each of the parallel resistors

Answer 1

R₁ = 100 Ohm;   R₄ = 1200 Ohm;      

  R₂ = 270 Ohm;   R₅ = 2700 Ohm;

  R₃ = 1000 Ohm; U_s = 12 V.

  I = ?; I_R3 = ?; I_R4 = ?; I_R5 = ?; U_R3R4R5 = ?; U_R1 = ?; U_R2 =?. 

1.     The value of the parallel combination of resistors R₃, R₄, R₅.

$R_345=(R_3* R_(4 )* R_5)/(R_4*R_5+R_3*R_5+R_3*R_(4 ) )= (1000*1200*2700)/(1200*2700+1000*2700+1200*1000)=453.7815 Ohm$

2.     The value of the total resistance of the circuit.

$R =R_1+R_2+R_345=100+270+453.7815=823.7815$ Ohm

3.     The supply current.

$I=U_s/R= 12/823.7815= 0.01456 A.$

4.     The voltage dropped R₁.

$U_R1=I*R_1= 0.01456*100=1.456 V.$

5.     The voltage dropped R₂.

$U_R2=I*R_2=0.01456*270=3.9312 V.$

6.     The voltage dropped R₃₄₅.

$U_345= U_s-U_R1-U_R2=12-1.456-3.9312=6.6128 V.$

7.     The current flowing through R_3.

$I_R3=U_345/R_3 =6.6128/1000=0.0066128 A.$

8.     The current flowing through R_4.

$I_R4=U_345/R_4 =6.6128/1200=0.0055106 A.$

9.     The current flowing through R_5.

$I_R5=U_345/R_5 =6.6128/2700=0.0024491 A.$