Four resistors are connected to a battery with a terminal voltageof 12 V, as sho

Question

Four resistors are connected to a battery with a terminal voltageof 12 V, as shown in the figure below. Use the following variablesas necessary: R1 = 26.0 ohm and R2 =79.0 ohm. (a) How would you reduce the circuit to anequivalent single resistor connected to the battery? Use thisprocedure to find the equivalent resistance of the circuit. 1correct check mark ohm (b) Find the current delivered by the battery to this equivalentresistance. 2correct check mark A (c) Determine the power delivered by the battery. 3correct check mark A (d) Determine the power delivered to the 50.0 ? resistor. 4wrong check mark Your answer is within 10% of the correctvalue. This may be due to roundoff error, or you could have amistake in your calculation. Carry out all intermediate results toat least four-digit accuracy to minimize roundoff error.W Part D I used the reasoningof how the current through the 50 ohm resistor is 0.2008A/2 =0.1004A so, power delivered = ((0.1004A)^2)*(50 ohm) = 0.5040 W I've tried to use 0.5, .5040, .5042 to no success. Only have onemore try left so I thought I'd ask before guessing .5041. Anysuggestions on if I'm using the wrong equation? Thanks!!

Explanation / Answer

. 1.   Req = 58.735   ;   yourresistance value is slightly high 2.   Ibattery = (12volts)/(58.735) = 204.3E-3ampere;   because your resistance is high,your current is low. 3.   Pbattery = (I)*(V) = (204.3E-3ampere)*(12 volts) = 2.452Watts 4.   I tried a slightly different approach;since I know the current through the 20 resistor, Icalculated the voltage across R2 which is the voltageacross all branches. I then used the voltage divider rule tocalculate the voltage across the 50 resistor. 5.   VR2 = (12.0 volts) - (204.3E-3ampere)*(20) = 7.914 volts 6.   V50 = (50/76)*(7.914volts) = 5.207 volts 7.   P50 =(V50)2/(50) = 542.3E-3Watts = 542.3 mW = 0.5423Watts .

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