For the circuit shown in the figure, R1 = 6.00 Ω, R2 = 6.00 Ω, R3 = 2.00 Ω, R4 = 4.00 Ω, R5 = 3.00 Ω, Physics fave answer!?

Question

For the circuit shown in the figure, R1 = 6.00 Ω, R2 = 6.00 Ω, R3 = 2.00 Ω, R4 = 4.00 Ω, R5 = 3.00 Ω, and the potential difference is 12.0 V.

aWhat is the equivalent resistance for the circuit?

bWhat is the current through R5?

cWhat is the potential drop across R3?

Philomel · Accepted Answer

1/22+4=6//6=3
6+3+3=12Ω
12/12=1A
a.   Rt=12Ω
b.   IR5=1A
IR3=1/2A
c.    ER3=1/2A*2Ω=1V

oubaas · Answer

Req = R1+ R2*(R3+R4) /(R2+R3+R4)+ R5 
Req = 6+6//6+3 = 6+6*6/12+3 = 6+3+3 = 12 ohm
I1 = I5 = V/Req = 12/12 = 1.0 A 
I3 = I1*R2/(R2+R3+R4) = 1.0*6/12 = 0.5 A 
V3 = R3*I3 = 2*0.5 = 1.0 V

az_lender · Answer

The combination of R2, R3, and R4 is equivalent to having a pair of 6-ohm resistors in parallel, so the resistance of this combination is 3 ohms.

The equivalent resistance in the whole circuit is
6 + 3 + 3 = 12 ohms.  The current through R5 is 1 amp.  The potential drop across R2 is 3 volts, and the potential drop across R3 is (2/6)*3V = 1 volt.

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For the circuit shown in the figure, R1 = 6.00 Ω, R2 = 6.00 Ω, R3 = 2.00 Ω, R4 = 4.00 Ω, R5 = 3.00 Ω, Physics fave answer!?

3 Answers