Four resistors are connected to a battery with a terminal voltage of 12 V, as sh

Question

Four resistors are connected to a battery with a terminal voltage of 12 V, as shown in the figure below. Use the following variables as necessary: R1 = 39.0 %u03A9 and R2 = 76.0 %u03A9.

(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 the circuit.

answer in Ohms

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

answer in A

(c) Determine the power delivered by the battery.

answer in W

(d) Determine the power delivered to the 50.0 %u03A9 resistor.

answer in W

Explanation / Answer

(a) R1 and the 50 ohm resistor are in series so the first equivalent resistor is:

Req1 = 50 + R1 = 50 + 39 = 89 Ohms

The first equivalent resistor and R2 are in parallel so their reciprocals add:

1/Req2 = 1/Req1 + 1/R2 = 1/89 + 1/76 = 0.02439

Req2 = 40.99 Ohms

finally, Req2 and the 20 Ohm resistor are in series so they add giving the total equivalent resistance.

Req = Req2 + 20 = 40.99 + 20 = 60.99 Ohms.

(b) V = IR => I = V/R = 12/60.99 = 0.1968 A

(c) P = IV = 0.1968*12 = 2.361 W

(d) the current going through Req is 0.1968 A. Since Req2 and the 20 Ohm resistor are in series they both have this same current. Thus the current running through Req2 is 0.1968 A. The voltage running through Req2 is: V = I*Req2 = 0.1968*40.99 = 8.067 V. Since R2 and Req1 are in parallel they have the same voltage as Req2. Thus the voltage through Req1 is 8.067 V. The current going through Req1 is the same as the current through the 50 ohm resistor and is: I = V/R = 8.067/89 = 0.09064 A.

Finally, the power in the 50 ohm resistor is:

P = I^2 R = (0.09064)^2*50 = 0.4108 W

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