# Electronics - Parallel Circuits - Discussion

Discussion Forum : Parallel Circuits - General Questions (Q.No. 28)

28.

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

Meter Readings: I = 10 mA, V = 12 V

Discussion:

8 comments Page 1 of 1.
Mohan rasu said:
8 years ago

Here, v=12v, r1=7.5 ohm, r2=5 ohm, r3=2 ohm.

WKT,

I=V/R.

I1=V/R1=12/7.5=1.6 mA.

I2=V/R2=12/5=2.4 mA.

I3=V/R3=12/2=6 mA.

Itot=10 mA.

So the ckt is operate normally.

WKT,

I=V/R.

I1=V/R1=12/7.5=1.6 mA.

I2=V/R2=12/5=2.4 mA.

I3=V/R3=12/2=6 mA.

Itot=10 mA.

So the ckt is operate normally.

Ash said:
9 years ago

How would you know that if r1 is open then c/n flow will be 8.4?

Waqar ahmad said:
9 years ago

Case 1.

When R1 is open then current flow in the circuit will be 8.4. Which is not equal to 10. So it is not correct.

Case 2.

When R2 is open then the current flow in the circuit will be 7.6 which is not correct as 10.

Case 3.

When fuse is open, the current and voltage would be zero.

Hence the circuit will operate normally.

When R1 is open then current flow in the circuit will be 8.4. Which is not equal to 10. So it is not correct.

Case 2.

When R2 is open then the current flow in the circuit will be 7.6 which is not correct as 10.

Case 3.

When fuse is open, the current and voltage would be zero.

Hence the circuit will operate normally.

Sunil Swarnkar said:
1 decade ago

Current through R1 = V/R1

= 12/7.5K = 1.6 mA

Current through R2 = V/R2

= 12/5K = 2.4 mA

Current through R3 = V/R3

= 12/2K = 6 mA

Total current 1.6+2.4+6 = 10mA

Ammeter is also showing 10mA.

So circuit is working normally.

= 12/7.5K = 1.6 mA

Current through R2 = V/R2

= 12/5K = 2.4 mA

Current through R3 = V/R3

= 12/2K = 6 mA

Total current 1.6+2.4+6 = 10mA

Ammeter is also showing 10mA.

So circuit is working normally.

Sreeyush Sudhakaran said:
1 decade ago

It=10mA

Vt=12v

Rt=Vt/It=12/10mA =1.2K

1/R1+1/R2+1/R3=1/1.2K

(1/7.5K)+(1/5K)+(1/2K)=0.833K (ALL K cancles)

0.133+0.2+0.5=0.833 is correct so none is open

circuit is operating normally

Vt=12v

Rt=Vt/It=12/10mA =1.2K

1/R1+1/R2+1/R3=1/1.2K

(1/7.5K)+(1/5K)+(1/2K)=0.833K (ALL K cancles)

0.133+0.2+0.5=0.833 is correct so none is open

circuit is operating normally

Url said:
1 decade ago

rt=0.00083

1/rt=1204.8129

after that,

it=vt/rt

so,

it=12/1204.8129

=9.96ma

1/rt=1204.8129

after that,

it=vt/rt

so,

it=12/1204.8129

=9.96ma

Teena said:
1 decade ago

Please give more explanation.

Naseem Ul Aziz said:
1 decade ago

Now we should know total current.

it=vt/rt

1/rt=(0.00013)+(0.0002)+(0.0005)=0.00083=1204.8196 ohm

it=9.9 ma

The circuit is operating normally

it=vt/rt

1/rt=(0.00013)+(0.0002)+(0.0005)=0.00083=1204.8196 ohm

it=9.9 ma

The circuit is operating normally

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