Solve the circuit shown in the picture by finding: (A) Equivalent Resistance. (B

Question

Solve the circuit shown in the picture by finding:

(A) Equivalent Resistance.

(B) Current Through EACH resistor.

(C) Voltage across EACH resistor.

(D) Power dissipated by EACH resistor.

Given:  Battery voltage = 12V    R1 = 5kOhm  R2= 5khm  R3 = 5kOhm  R4=5kOhm

[if the picture is not shown, R1 & R2 are in series and R3 & R4 are parallel)

Solve the circuit shown in the picture by finding: Equivalent Resistance. Current Through EACH resistor. Voltage across EACH resistor. Power dissipated by EACH resistor. Given: Battery voltage = 12V R1 = 5kOhm R2= 5khm R3 = 5kOhm R4=5kOhm [if the picture is not shown, R1 & R2 are in series and R3 & R4 are parallel)

Explanation / Answer

a)equivalent resistance

R3 and R4 are in parallel

R34 =R3*R4/(R3+R4) =5*5/(5+5)R34 =2.5 Kohm

R1 ,R2 and R34 are in series

Req=R1+R2+R34 =5+5+2.5

Req =12.5 Kohm

b)total current

I =V/R =12/12.5*10^3

I =0.96*10^-3 A or 0.96 mA

so

I1 =I2 =I34 =0.96 mA or 9.6*10^-4 A

I3 =I4 =0.96/2 =0.48 mA or 4.8*10^-4 A

c)

V1 =I1*R1 =9.6*10^-4*5*10^3 =4.8 V

V2 =I2R2 =9.6*10^-4*5*10^3 =4.8 V

V3=I3R3=4.8*10^-4*5*10^3 =2.4 V

V4 =I4R4 =2.4 V

d)

P1 =V1I1 =4.61*10^-3 W or 4.61 mW

P2=V2I2 =4.61*10^-3 W or 4.61 mW

P3 =V3I3 =2.4*4.8*10^-4 =1.15*10^-3 or 1.15 mA

P4=V4I4 =1.15*10^-3 or 1.15 mA

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