Question

Four resistorsare connected to a battery with a terminal voltage of 12 V, as shown in the figure below. Use the following variables as necessary: R1 = 34.0 Ω and R2 = 85.0Ω.

(a) How would you reduce the circuit to an equivalent single resistor connected to the battery? Use this procedure to find the equivalent resistance of thecircuit.
________ Ω

(b) Find the current delivered by the battery to this equivalent resistance.
________ A

(c) Determine the power delivered by the battery.
______ W

(d) Determine the power delivered to the 50.0 Ω resistor.
_________ W

Answer 1

fr equivalent resistance
first
R1 and 50 are in series there fore Req1= 50 +34=84
now Req1 and R2 are parallel = Req2 = Req1*R2/(Req2+R2) = 42.248 ohms
now Req2 and 20 are in series
this implies R eq circuit = 62.248 ohms connetceted to battery of 12 volts
I= V/r = 0.284 A
power delivered = V²/ R = 2.313 w

power delivered to 50 ohm resister =

I through 50 ohms = I(84/(85+84)) = 0.1232

power = I²R = 0.76 w

Answer 2

Req=1/(1/(34+50)+1/85)+20≈62.249 Ω==>12 V/62.249≈0.193 A==>I²*R

0.193²*62.249≈2.313 W==>V_R20=I*R=193 mA*20 Ω≈3.86 V∴12 V-3.86 V=8.14 V

across the parallel circuits==>8.14/84≈97 mA==>P₅₀=I²*R=0.097²*50≈0.47 W

a) 62.25 Ω

b) 0.193 A

c) 2.313 W

d) 0.47 W