Electrical Engineering - Series Circuits - Discussion

@All.

The question is a little tricky.

1) If we consider it as a series circuit with very very little current (very high resistance except for R3) say in micro amps then only it's possible to get nearly 6v across R3 (not exactly 6v).

2) In case if the resistors except R3 got damaged (shorted) then only 6v across R3.

The series circuit current is same voltage is divided according to statement all resistor voltage drop is zero then voltage appears r3 resistor so r3=6v.

We know in series,

V=V1+V2+V3+V4+V5
=>6=0+0+V3+0+0
=>V3=6v.

R3 is load.

If all resistors (except R3 are) considered having zero ohm resistance, then this condition satisfies.

R3 is an open circuit, so all other resistor voltage is 0, r3 will show 6 voltage.

Yes, you are correct @Abdul.

V = IR.
R = V/I,
V = I*V/I.
V = V=6V.

It is the correct answer.

It is 6v.

Question states we have voltage to the circuit and Its equal to 6v. (No dead batteries ect. Sorry)

ALL EXCEPT r3 have zero voltage drop-- the question is stating r3 has a nonzero voltage drop. If all the other resistors have zero drops they have zero resistance-- like ideal wire.

So we only have a single resistor in the circuit r3.

If these resistors were in parallel you wouldn't have a circuit for long because you'd have 4 shorts. The question states a series circuit.

So, the voltage drop across a single resistor in a 6v circuit. Then the answer is 6v. No math required.

The answer is 6v.

> voltage is zero when current is zero or resistance is zero.
Current cannot be zero.
So resistance is zero, may be a short circuit across the resistors.