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?

Discussion:
23 comments Page 3 of 3.
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)
Shahid said:
4 years ago
V/R =6/300 =0.02A
Load time = Battery Capacity in Ah/Load Current.
= 40Ah/.02A = 2000h.
Load time = Battery Capacity in Ah/Load Current.
= 40Ah/.02A = 2000h.
(7)
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.
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