Electronics - Series-Parallel Circuits - Discussion

Discussion Forum : Series-Parallel Circuits - General Questions (Q.No. 3)

Series-Parallel Circuits

If the load in the given circuit is 120 k , what is the loaded output voltage?

4.21 V

15.79 V

16 V

19.67 V

Answer: Option

Explanation:

No answer description is available. Let's discuss.

Discussion:

19 comments Page 1 of 2.

AKHIL VA said: 6 years ago

resistor 120k is parallel with 8k.
So r=1/120+1/8 = 7.5k.
vout=r2/(r1+r2) = 7.5/((2+7.5).
= 15.78947368421053 V,
= 15.79 V.

(5)

Alex said: 7 years ago

Is this series, parallel, or an open circuit? Please tell me.

(2)

Thiru said: 8 years ago

Can we choose this method?

Apply *voltage Divider Rule *
V=((Total voltage *opposite Resistance) /Total Resistance)
20*8/8+2=16V.

Whatever the Voltage across parallel is same as the other side That's why its 16v.

Vc_davv said: 8 years ago

Don't be so complex.

Use thevenin theorem.
So vth = 16 v and Rth = 1.6 kohm.
Voltage across Rl (load 120kohm) by VDL = (16*120K/120k+1.6k)=15.7v.

(1)

Shik shansha vali said: 8 years ago

RL = 120 anyone can explain?

Pragadee said: 8 years ago

Thanks for explaining this.

Vaishnavi said: 8 years ago

Thanks for explaining it.

Aravind said: 8 years ago

8k parallel 10k= (8*120)/(8+120)= 7.5k.

Now,

Voltage across 7.5k = Vol.across 120k (bz 8k parallel 120k)

So vol across 7.5k =(20V * 7.5k)/2k + 7.5k,
= 15.79V.

(Vol.divider equation)

(5)

Yoosuf said: 9 years ago

Thanks for explaining the solution.

Manoj said: 9 years ago

Thanks @Rekha.

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