Question

In the following circuit, the battery emf is 25V, R1 = 3 kN, R2 = 5 kN, R3 = 1.5 kN, R4 = 8 kN, R5 = 1 kN, Ro = 2 k12, -R2 R4 .Ro w R1 R3 RE What current is drawn from the battery in mA?

b) What current flows through R6 in mA?

Answer 1

given data

R₁ =3KΩ ,  R₂ =5KΩ , R₃ = 1.5KΩ, R₄= 8KΩ ,R₅ = 1KΩ ,R₆ =2KΩ

EMF E=V = 25V

in this circuit

R₄ & R₆ are in parallel combination

1/Req₁ =1/R₄ +1/R₆

1/Req₁ = 1/8KΩ +1/2KΩ = 5/8KΩ

R_eq1 = (8/5) KΩ

R_eq1 and R₅  are in series then

R_eq2 =   R_eq1 +R₅ = (8/5) KΩ +1KΩ = (13/5)KΩ

R₂ and R₃  are in series then

R_eq3= R₂+R₃ =5KΩ+1.5KΩ = 6.5KΩ

R₁, R_eq2 and R_eq3 are in parallel then

1/R_EQ = 1/R₁ +1/ R_eq2+1/R_eq3

1/R_EQ =( 1/3KΩ )+( 1/(13/5)KΩ ) +( 1/ 6.5KΩ ) =1/0.8718

R_EQ = 1.147 KΩ = 1147 KΩ

CURRENT DRAWN IN THE Circuit

V = I R_EQ =  

I = V/R_EQ = 25/1147 = 21.795 x10^-3A = 21.795 mA

I = 21.795 mA = 21.795 mA

CURRENT DRAWN IN THE Circuit I = 21.795 mA

B)

Current through R₅ is I₅ = V/R_eq2 = 25/ (13/5) kΩ = 9.615 mA

I₅ =9.615X10^-3 A

Current through resistor R₆ is I₆ = (R₄ /(R₄+R₆) I₅ = (8KΩ /(8KΩ +2KΩ ))(9.615X10^-3 A) = 7.692 X10^-3 A

I₆ = 7.692 mA

Current through resistor R₆ is I₆ = 7.692 mA

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