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
. The percentage of the total voltage appearing across a 47 resistor that makes up part of the total series resistance isDiscussion:
13 comments Page 1 of 2.
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%.
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%.
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%.
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%.
Jagadesh said:
9 years ago
The percentage of voltage across any resistor = (Resistor value/Total resistance)*100.
Percentage v = (47/680)*100.
= 6.9%.
Percentage v = (47/680)*100.
= 6.9%.
(1)
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%.
Then at 47-ohm resistance = Voltage is?
47 *100/680 = 6.9117647045 ~ 6.91%.
(1)
Ali Alrushoud said:
1 decade ago
I constant in series circuit
V=IR = I 680
V'=Ir = I 47
V'/V = I 47 / I 680 = 47 / 680 = 0.0691
V' = 6.91% V
V=IR = I 680
V'=Ir = I 47
V'/V = I 47 / I 680 = 47 / 680 = 0.0691
V' = 6.91% V
DILEEP M JOHN said:
1 decade ago
V=IR
series----I constant
v prop. R
v prop 680
v' prop 47
dividing v'=47/680V
v'=.069V
ie 6.9 %
series----I constant
v prop. R
v prop 680
v' prop 47
dividing v'=47/680V
v'=.069V
ie 6.9 %
The BOSS said:
1 decade ago
100%v at 680ohm
-(x)-%v at 47ohm
find x (cross multiplication)
x = 100.47 / 680 = 6.91%
-(x)-%v at 47ohm
find x (cross multiplication)
x = 100.47 / 680 = 6.91%
VIDYA SAGAR said:
1 decade ago
680....>100%.
47....>?
680*X = 47*100.
X = 4700/680.
X = 6.91%.
47....>?
680*X = 47*100.
X = 4700/680.
X = 6.91%.
Allulavu said:
3 months ago
Thanks all for explaining the answer.
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