Find the potential difference across each resistor in the figure below. (R1 4.60

Question

Find the potential difference across each resistor in the figure below. (R1 4.60 o, R2 = 4.40 n, R3 = 2.70a, R4 = 2.25 o) 12.0 Need Help? Show My Work (Optional) For parts (a), (b), and (c) use the figure below. (Use the following as necessary: v = 15.0 v, R1 = 5.00 , and R2 = 10.00 .) (a) Is it possible to reduce the circuit shown in the figure above to a single equivalent resistor connected across the battery? Explain. (b) Find the current in the 5.00 resistor. (c) Calculate the power delivered by the battery to the circuit. Need Help? Show My Work (Optional)

Explanation / Answer

ans. 7) a) 3 ohm and 10 ohm resistance in parallel will equivallent to Req1 = 3*10/(3+10) = 2.307 ohm

now 5 ohm , Req1 , 4 ohm are in series will equivallent to Req = 5+2.307+4 = 11.307 ohm

=> Req = 11.307 ohm single eq. resistance connected across the battery.

b) current through the battery

i = V/Req = 15/11.307 = 1.3266 A = current through 5 ohm resistor.

c) power delivered by the battery

P = VI = 15*1.3266 = 19.899 watt

ans. 6) let voltage at upper junction is V and at lower junction is zero(ground potential)

i1,i2,i3 are currents in the branches R1, R2,R3 respectively

apply KCL at upper junction

i1+i2+i3 = 0

=> [0 - (V-12)]/4.60 + [ 0 - (V-3)]/4.40 + [0 - (V - 18)]/(2.70+2.25) = 0

=> (12-V)/4.60 + (3-V)/4.40 + (18-V)/4.95 = 0

=> 2.608 - 0.217V + 0.681 - 0.227V + 3.636 - 0.202V = 0

=> V = 10.719 volt

SO V(R1) = 0 - (V-12) = 12 - V = 1.281 volt

V(R2) = 0 -(V-3) = 3 - V = - 7.719 volt

V(R3 + R4) = 0 - (V - 18) = 18 - V = 7.281 volt

=> V(R3) = 7.281*2.70/(4.95) = 3.97 volt

=> V(R4) = 7.281 - 3.97 = 3.311 volt  

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