Electrical Engineering - Parallel Circuits - Discussion

Discussion Forum : Parallel Circuits - General Questions (Q.No. 20)
20.
Three 47 resistors are connected in parallel across a 110 volt source. The current drawn from the source is approximately
2.3 A
780 mA
47 mA
7.06 A
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Anomie said:   5 years ago
(47^-1*3)^-1 = 15.666.
110/15.666 = 7.02A.

Poovarasan said:   8 years ago
If the resistor is connected in series across an 110-volt source.

Raj said:   8 years ago
I tried the calculation in calci, it gives 100/15.6 is 6.38. I am confused, please help me.

Pritam said:   8 years ago
It should be 7.02 amp.

I = 110*3/47 = 7.0212 amp.
(2)

Heartless boy (omi) said:   1 decade ago
Req- 47/3.
= 15.67.

Now,
I = V/R.

= 110/15.67.
= 7.06 amp.
(2)

Debaraj sethi said:   1 decade ago
Parallel in resistance is same value r/n = 47/3 = 15.66.

Rt = 15.66.

V= 110.

I = 110/15.66 = 7.06v.

V = IR,
V = 110,
R = 15.6{Parallel in resistance so, r = 1/r1+1/r2+1/r3(3 phase)}.

I = V/R = 110/15.6 = 7.06V.

Given 3 resistances are connected in parallel.
Each resistance value (R) = 47 ohm.
Two resistance value is 23.5.
Resistance 23.5 is parallel to another 45ohm resistor.
The value of (Req) =15. 66.
We can know formulla I=V/R.
The value of (I) is=7.02A The approximate answer given problem is 7.6.
(1)

Mukesh Kumar said:   1 decade ago
1/R=1/R1+1/R2+1/R3=1/47+1/47+1/47=3/47
R=47/3=15.67

I=V/R=110/15.67=7.06A
(3)

Equation should be
Req=R/N
Req=R1/1+R2/1+R3/1
Req=47/3
Req=15.6 ohms
I=V/R
I=110/15.6
I=7.06 A
(3)