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.
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)
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)
Rajakumar said:
6 years ago
Current through load I=V/R.
I=6/300=0.02 then,
Rating=40AH,
40=0.02H.
So, Duration H=2000.
I=6/300=0.02 then,
Rating=40AH,
40=0.02H.
So, Duration H=2000.
(1)
Shweta said:
1 decade ago
I = V/R.
= 6/300.
= 0.02A.
Ah = 40.
H = 40/A.
= 40/0.02.
= 2000.
= 6/300.
= 0.02A.
Ah = 40.
H = 40/A.
= 40/0.02.
= 2000.
(1)
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.
Suri said:
7 years ago
V=I x R.
6V=I x 300ohms.
I=6V / 300ohms.
I=0.02Amps.
40Amps-hour / 0.02Amps = 2000 hours.
2000 Hours Answer.
6V=I x 300ohms.
I=6V / 300ohms.
I=0.02Amps.
40Amps-hour / 0.02Amps = 2000 hours.
2000 Hours Answer.
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.
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.
Padma said:
8 years ago
Yes, you are Correct @Sabir.
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