Electrical Engineering - Series Circuits - Discussion

Discussion Forum : Series Circuits - General Questions (Q.No. 24)
24.
The total resistance of a circuit is 680 . The percentage of the total voltage appearing across a 47 resistor that makes up part of the total series resistance is
68%
47%
69%
6.91%
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Allulavu said:   3 months ago
Thanks all for explaining the answer.

TUSHAR NAKUM said:   3 years ago
At 680 ohm Resistance = Voltage is 100%,
Then at 47-ohm resistance = Voltage is?
47 *100/680 = 6.9117647045 ~ 6.91%.
(1)

Karthik Netha said:   4 years ago
@Boss.

Nice explanation Thanks.

Sandy said:   9 years ago
Awesome explanation @Sandeep.

Jagadesh said:   9 years ago
The percentage of voltage across any resistor = (Resistor value/Total resistance)*100.

Percentage v = (47/680)*100.

= 6.9%.
(1)

Sandeep patil said:   10 years ago
Above all answers are correct.

Let me answer in different way. Since total resistance is 680 ohms. One of them is 47 ohm.

So 680-47 = 633 ohms right, means will imagine one resistor is having 633 ohms and another is 47 ohms ok.

And in above options which option is having least % that will be the % voltage across 47 ohm resistor. i.e. 6.9%.

Mahesh said:   1 decade ago
Since it is asked in percentage lets take V as 100%.

Total series resistance is 680 ohms.

For 680 it takes 100% of V.

680 ohms =100%.
47 ohms = ?

Cross multiply you will get 6.91%.

Drethi said:   1 decade ago
I didn't get. Please explain me.

Sairam said:   1 decade ago
Since voltage and resistance are having linear relationship so % of voltage drop across 47 Ohm resistance can be calculated as 47/680 = 6.91%.

VIDYA SAGAR said:   1 decade ago
680....>100%.

47....>?

680*X = 47*100.

X = 4700/680.

X = 6.91%.


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