# Electronics - Parallel Circuits - Discussion

Discussion Forum : Parallel Circuits - General Questions (Q.No. 11)

11.

What would these meter readings indicate about the circuit in the given circuit?

Meter Readings: I = 7.6 mA, V = 12 V

Discussion:

17 comments Page 1 of 2.
Himanshu Verma said:
1 decade ago

Case1 :When R1 is open,

then the effective resistance R = R2||R3

i.e. 5K||2K = 10/7 K

then current I = V/R = (12*7)/10K = 8.4 mA(Which is not correct ans i.e 7.6 mA)

Case2 : When R2 is open,

then the effective resistance R = R1||R3

i.e. 7.5K||2K = 30/19 K

then current I = V/R = (12*19)/30K = 7.6 mA(Which is d correct ans i.e 7.6 mA)

Case3: When the fuse is open,

then it is not possible to obtain voltage of 12 V

So the correct answer is option [b].

then the effective resistance R = R2||R3

i.e. 5K||2K = 10/7 K

then current I = V/R = (12*7)/10K = 8.4 mA(Which is not correct ans i.e 7.6 mA)

Case2 : When R2 is open,

then the effective resistance R = R1||R3

i.e. 7.5K||2K = 30/19 K

then current I = V/R = (12*19)/30K = 7.6 mA(Which is d correct ans i.e 7.6 mA)

Case3: When the fuse is open,

then it is not possible to obtain voltage of 12 V

So the correct answer is option [b].

Saichaitanya said:
1 decade ago

CONSIDER THE FOLLOWING:

In Parallel circuit, the current will differ and the voltage will remain same across all the branches. In other words, the voltage across R1, R2 and R3 will be 12V each.

I1= 12/7.5K => 1.6mA

I2= 12/5K => 2.4mA

I3= 12/2K => 6.0mA

In above, the total current 7.6mA will take place correctly when only I1 and I3 present *(since, I1 + I3 = 7.6mA).

Hence, R2 must be discarded/ remove from the circuit. In other words, R2 is open.

In Parallel circuit, the current will differ and the voltage will remain same across all the branches. In other words, the voltage across R1, R2 and R3 will be 12V each.

I1= 12/7.5K => 1.6mA

I2= 12/5K => 2.4mA

I3= 12/2K => 6.0mA

In above, the total current 7.6mA will take place correctly when only I1 and I3 present *(since, I1 + I3 = 7.6mA).

Hence, R2 must be discarded/ remove from the circuit. In other words, R2 is open.

(1)

Mohan said:
1 decade ago

I agree with Saichaithanya.

Sudhansu said:
1 decade ago

Well done Saichaitanya and Himanshu Verma.

Shinoj said:
1 decade ago

@ Saichaitanya, Perfect explanation.

Sriram pn said:
1 decade ago

Saichaitanya perfect explanation

Suthan said:
1 decade ago

Fantastic Himanshu Verma & Saichaitanya ..Great work.

Gigi said:
1 decade ago

Do the total resistance in the cut by V/I that is 1.578.

Then do the total combination or resist ivory in parallel which resisters combination gives 1.578 ohms and other resistor is open.

Then do the total combination or resist ivory in parallel which resisters combination gives 1.578 ohms and other resistor is open.

Santosh said:
1 decade ago

Voltmeter is in parallel connection, so reading is 12V.

Now for resistance calculation which is connected parallel with 2k ohm resistance (R3):

Equivalent resis. R = 12V/7.6mA = (30/19)K ohm.

So, Rx ll R3 = (30/19)k.

or, (1/Rx)+ (1+R3) = 1/(30/19)k.

or, (1/Rx)+ (1/2) = (19/30).

or, (1/Rx) = (19/30)-(1/2).

or, 1/Rx = 4/30.

or, Rx = 30/4 = 7.5k ohm.

i.e., shows R1 is connected & R2 is opened.

Now for resistance calculation which is connected parallel with 2k ohm resistance (R3):

Equivalent resis. R = 12V/7.6mA = (30/19)K ohm.

So, Rx ll R3 = (30/19)k.

or, (1/Rx)+ (1+R3) = 1/(30/19)k.

or, (1/Rx)+ (1/2) = (19/30).

or, (1/Rx) = (19/30)-(1/2).

or, 1/Rx = 4/30.

or, Rx = 30/4 = 7.5k ohm.

i.e., shows R1 is connected & R2 is opened.

(1)

Pre said:
8 years ago

Well done @ Himanshu Verma.

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