Electrical Engineering - Energy and Power - Discussion
Discussion Forum : Energy and Power - General Questions (Q.No. 11)
11.
A 6 V battery is connected to a 300 
load. Under these conditions, it is rated at 40 Ah. How long can it supply current to the load?
load. Under these conditions, it is rated at 40 Ah. How long can it supply current to the load?Discussion:
23 comments Page 1 of 3.
SIRKUDOS said:
8 years ago
First get the power consumed.
V=IR I = V/R =6/300 = 0.02A.
P = IV = 0.02*6=0.12W. That's is the load power.
To find the battery power.
P=IV 6*40 = 240Wh.
Then to find the hour the load will run.
240wh/0.12w = 2000h.
V=IR I = V/R =6/300 = 0.02A.
P = IV = 0.02*6=0.12W. That's is the load power.
To find the battery power.
P=IV 6*40 = 240Wh.
Then to find the hour the load will run.
240wh/0.12w = 2000h.
Upendra said:
1 decade ago
Actually current taken by 300ohm load is 6/300=0.02Amp. If condition saying,battery is under 40AH then to cover 40Amps, load has to take 2000hr at a constant current of 0.02Amps.
So answer is 'C'.
So answer is 'C'.
T.T.ACHARYA said:
1 decade ago
The procedure is as follows:
=> ampere(A)xhour(T)=(voltage(V)/resistance(R))xhour(T)
=> 40 =(6/300) X T
=> T =(40X300)/6
=> T =2000h
=> ampere(A)xhour(T)=(voltage(V)/resistance(R))xhour(T)
=> 40 =(6/300) X T
=> T =(40X300)/6
=> T =2000h
SASMITA MAHALIK said:
2 weeks ago
Simply it can solve by this method:.
It's given : Ah=40 so, h=40/A.
TO find A=?
I=V/R, = 6/300=0. 02A. Now put in above 'h' eq.
H=40/A, = 40/0.02 = 2000 hours.
It's given : Ah=40 so, h=40/A.
TO find A=?
I=V/R, = 6/300=0. 02A. Now put in above 'h' eq.
H=40/A, = 40/0.02 = 2000 hours.
HAris said:
6 years ago
Simply it can solve by this method:
It's given : Ah=40 so, h=40/A.
TO find A=?
I=V/R, = 6/300=0.02A. Now put in above 'h' eq.
h=40/A, = 40/0.02= 2000 hours.
It's given : Ah=40 so, h=40/A.
TO find A=?
I=V/R, = 6/300=0.02A. Now put in above 'h' eq.
h=40/A, = 40/0.02= 2000 hours.
(4)
Phani bhargav.v said:
1 decade ago
Actually load may be or may not be estimate like this as 300 ohms, why because power consumption is depend upon their capacity in fact their wattage.
Md. Muin Uddin said:
1 decade ago
Energy produce by battery = 40*6 = 240whr
Power taken by load = v*v/r = 0.12
So current supplying time = 240/.12=2000hr.
Power taken by load = v*v/r = 0.12
So current supplying time = 240/.12=2000hr.
Ashraf said:
1 decade ago
Energy produce by battery = 40*6 = 240whr
Power taken by load = v*v/r = 0.12
So current supplying time = 240/.12=2000hr
Power taken by load = v*v/r = 0.12
So current supplying time = 240/.12=2000hr
Rajajee said:
1 decade ago
In the above problem given ah=40 and we get i from v=i*r as 0.02.
So in ah=40 amp a=0.02, So .02*h=40.
h=40/0.02=2000.
So in ah=40 amp a=0.02, So .02*h=40.
h=40/0.02=2000.
Parth said:
8 years ago
I = V/R.
= 6/300,
= 0.02.
As we know it rates 40Ah, thus,
40 = I * t.
40 = 0.02 * t,
t = 40/0.02,
t = 2000 h.
= 6/300,
= 0.02.
As we know it rates 40Ah, thus,
40 = I * t.
40 = 0.02 * t,
t = 40/0.02,
t = 2000 h.
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