Question

zka or the circuit shown in the figure, a) Calculate the currents in each resistor.--て 12 Calculate the voltage across each resistor. V c) Calculate the power dissipation in each resistor.279 d) Calculate the power output of the battery.How does it compare to your results in part (c) b)

Answer 1

First of all we have to find equivalent resistance
R₂ and R₃ are in parallel , therefore it's equivalent resistance
(1/R) = (1/R₂)+ (1/R₃) = (1/2)+(1/3)
R = 1.2 k$\Omega$
Now this will be in the series with R₁ , hence the net resistance of the circuit
R_Net = 4 + 1.2 = 5.2 k$\Omega$
(a) Current in the circuit
i = V/R_Net = 12/(5.2*10³)= 2.308*10^-3 A
Hence the current in the resistor R₁ is 2.308 mA
Potential drop in the resistor = iR = 2.308*10^-3*4*10³ = 9.232 V
Power dissipiation in the R₁ = Vi = 9.232*2.308*10^-3 = 21.307 *10^-3 W
Now the potential drop in the R₂ and R₃ will be same becasue they are in parallel and it will
equal to = 12 - 9.232 = 2.768 V
Voltage in resistance R₂ , = 2.768 V
Current in resistance R_{2 ,} i₁ = V/R = 2.768 / (2*10³) = 1.384*10^-3 A
Power in the resistance R₂ = vi₁ = 2.768*1.384*10^-3 = 3.831*10^-3 W
Now in resistance R₃
Voltage = 2.768 V
i₂ = 2.768 /(3*10³) = 0.923*10^-3 A
Power = vi = 2.768*0.923*10^-3 = 2.55*10^-3 W
Power output of the battery = Vi = 12*2.308*10^-3 = 27.696*10^-3 W
When we add power dissipiation of all resistor it will equal to the power of the battery.

12 12v