Electronics - Series-Parallel Circuits - Discussion

Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 34)
34.
With 6 V applied, what is the voltage across R2 of a parallel combination of R2 (1 k omega.gif), R3 (2 k omega.gif), and R4 (1 k omega.gif) in series with R1, whose value is 2 k omega.gif?
1 V
3 V
5 V
6 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
8 comments Page 1 of 1.

Saumitra said:   1 decade ago
1st-R1 series with (parallel combn of R2, R3, R4).

Overall resistance is 2.4 ohm i.e. 2+0.4.

Now make convert 6v to 3amp and R1 (2 ohm) parallel to 3 amp (i.e voltage to current conversation).

Apply current division rule you got I2 A = 0.5 amp (current through 2 ohm resistor).

Then v = IR, V = 0.5*2 = 1 v.

(please confirm again, still I have doubt).

KIRAN V said:   9 years ago
First find the Total Resistance

R2 || R3 || R4 = 0.4 K Ohm.
Rtotal = (R1 + (R2 || R3 || R4) ) = 2 + 0.4 = 2.4 K Ohm.

Now use Voltage Divider formula,
V1 = Vin * (R1 / (R1 + (R2 || R3 || R4))).
V1 = 6 * (2 * 10^3/ (2.4 * 10^3)),
V1 = 5 V,
V = V1 + V2.
V = 5 + V2,
V2 = V - 5,
V2 = 1 V.

Voltage is same at every point in Parallel combination.

Nan_mona said:   8 years ago
R2=R2||R3||R4
=2/4 K ohm.
Total resistance = R1+R2
= 2 + 2/4 = 10/4 K ohm.

Using potential divider rule,
o/p V = (total vol x opp branch resistance)/total resistance.
V = (6*(2/4))/(10/4).
= 1 V.

Shivany said:   1 decade ago
R2, R3, R4 are in parallel and R1 is in series with all three right? Then it would be 2.4 kohm I guess. How 2.57?

RAJAT said:   1 decade ago
Rtotal=2.57ohm
Itotal=v/r=6/2.57=2.33A
by voltage dividerrule
voltage accross R2=6*1/2+1+2+1=1V.

Poojitha said:   9 years ago
Please explain the answer step by step by writing the formula also.

Pratheep said:   1 decade ago
The answer suppose to be 1.636V.

MONA_NAN said:   8 years ago
The answer suppose to be 1.2 V.

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