Electrical Engineering - Parallel Circuits - Discussion
Discussion Forum : Parallel Circuits - General Questions (Q.No. 22)
22.
Four resistors are connected in parallel. Fifteen mA flows through resistor R. If the second resistor is 2R, the third resistor 3R, and the fourth resistor 4R, the total current in the circuit is
Discussion:
6 comments Page 1 of 1.
Maybe said:
8 years ago
Since voltage remains constant.
branch 1->V=IR.
branch 2->V=(I/2)2R.
So on
thus Itotal=15+(15/2)+(15/3)+(15/4)=31.25mA.
branch 1->V=IR.
branch 2->V=(I/2)2R.
So on
thus Itotal=15+(15/2)+(15/3)+(15/4)=31.25mA.
(1)
Maybe said:
8 years ago
Since voltage remains constant,
branch 1->V=IR,
branch 2->V=(I/2)2R,
So on.
Thus Itotal=15+(15/2)+(15/3)+(15/4)=31.25mA.
branch 1->V=IR,
branch 2->V=(I/2)2R,
So on.
Thus Itotal=15+(15/2)+(15/3)+(15/4)=31.25mA.
Kichu said:
8 years ago
Well said @Twumasi.
Twumasi samuel said:
1 decade ago
15*R = I*2R
I = 15/2 = 7.5.
= 15/3 = 5.
= 15/4 = 3.75.
TOTAL current is 15+7.5+5+3.75 = 31.25.
I = 15/2 = 7.5.
= 15/3 = 5.
= 15/4 = 3.75.
TOTAL current is 15+7.5+5+3.75 = 31.25.
(3)
Sreeraj said:
1 decade ago
When resistance increases current decrease.
First way of solving
---------------------
I1=15--->R
I2=15/2---->2R
I3=15/3----->3R
I4=15/4------>4R
I=I1+I2+I3+I4=31.25
Second way of solving
---------------------
I1=(It*Rt)/R----(1)
It-->Total current
Rt-->Total resistance
It*Rt---->Voltage
Rt=1/(1/R+1/2R+1/3R+1/4R)=12R/25
From (1) equation
15=(It*12R/25)/R=It*12R/25*R=It*12/25
=>It=15*25/12=31.25
First way of solving
---------------------
I1=15--->R
I2=15/2---->2R
I3=15/3----->3R
I4=15/4------>4R
I=I1+I2+I3+I4=31.25
Second way of solving
---------------------
I1=(It*Rt)/R----(1)
It-->Total current
Rt-->Total resistance
It*Rt---->Voltage
Rt=1/(1/R+1/2R+1/3R+1/4R)=12R/25
From (1) equation
15=(It*12R/25)/R=It*12R/25*R=It*12/25
=>It=15*25/12=31.25
(2)
Mohana said:
1 decade ago
Someone please get me the explanation.
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