Electrical Engineering - Energy and Power - Discussion
Discussion Forum : Energy and Power - General Questions (Q.No. 7)
7.
A 15 V source is connected across a 12 
resistor. How much energy is used in three minutes?
resistor. How much energy is used in three minutes?Discussion:
31 comments Page 1 of 4.
Fakhur said:
2 years ago
I will assume that this is not in a very hot ambient temperature nor is it sandwiched between other hot components.
In that case, good engineering practice is to use a resistor rated for at least 2x the required power dissipation.
One formula for power is I squared times R.
P = 0.025A^2 x 120 ohms = 0.075W or 75mW.
So, a resistor rated at least double that 150mW.
In that case, good engineering practice is to use a resistor rated for at least 2x the required power dissipation.
One formula for power is I squared times R.
P = 0.025A^2 x 120 ohms = 0.075W or 75mW.
So, a resistor rated at least double that 150mW.
(3)
Dhina said:
4 years ago
Energy=power *Time
J=w*s
V=15
3mins/60 = 0.05 s
R=12.
Power=I*v
I=V/R
=15/12.
I=1.25.
P=1.25*5,
P=18.75.
J= 18.75 * 0.05.
J= 0.9375.
J=w*s
V=15
3mins/60 = 0.05 s
R=12.
Power=I*v
I=V/R
=15/12.
I=1.25.
P=1.25*5,
P=18.75.
J= 18.75 * 0.05.
J= 0.9375.
(12)
Dhanajayaalthi said:
5 years ago
Voltage= 15v.
Resistor= 12 ohms.
POWER(P) = VI * t.
= (V*(V/R) ) * t Bcoz I= V/R
= ((V)^2 /R) *t
= ((15)^2 /12 )*(3/60) (for 3 mins)
= (225/12) * (1/20)
= 18.76 * 0.05
= 0.938 kwh.
Resistor= 12 ohms.
POWER(P) = VI * t.
= (V*(V/R) ) * t Bcoz I= V/R
= ((V)^2 /R) *t
= ((15)^2 /12 )*(3/60) (for 3 mins)
= (225/12) * (1/20)
= 18.76 * 0.05
= 0.938 kwh.
(4)
Gaikwad.pravin said:
6 years ago
Here, it is (v^2/R) *3.
BhuPavKar said:
6 years ago
WKT, P=[(I)^2] * R (Unit - Watts) --> eqn 1.
Energy = E = P * t (Unit - Watt-hour) --> eqn 2.
Given = t = 3 min; But, t in hours = (3/60) hrs.
Next, I = V/R = 15/12 = 1.25.
Sub t and I in eqn 2, we get E = 0.9375Wh.
Energy = E = P * t (Unit - Watt-hour) --> eqn 2.
Given = t = 3 min; But, t in hours = (3/60) hrs.
Next, I = V/R = 15/12 = 1.25.
Sub t and I in eqn 2, we get E = 0.9375Wh.
(4)
Bibek said:
7 years ago
we know that, E=i*i* R *T.
I=V/R =15/12=1.25 A,
T=3 MIN,
60 MIN=1 HOUR,
3MIN=3/60 HOUR=.05 H.
E=1.25*1.25*12*.05=0.9375 Wh.
I=V/R =15/12=1.25 A,
T=3 MIN,
60 MIN=1 HOUR,
3MIN=3/60 HOUR=.05 H.
E=1.25*1.25*12*.05=0.9375 Wh.
Kgabisang said:
7 years ago
v=IR
I=V/R
=15V/12ohm
=1.25A.
P=VI
=15V * 1.25A
= 18.5W.
Energy = Power * time
=18.75W * 0.05h
=0.938 Wh.
I=V/R
=15V/12ohm
=1.25A.
P=VI
=15V * 1.25A
= 18.5W.
Energy = Power * time
=18.75W * 0.05h
=0.938 Wh.
(4)
Satish mahadu kukade said:
8 years ago
You are correct @Debu.
Zahid said:
8 years ago
Well said @Bains and @Debu.
Minnu said:
8 years ago
P = V^2/R = (15)^2/12,
= 225/12 = 18.75,
t = 3 minutes = (3/60) Hours = 0.05 Hours,
So, ENERGY consumption in 3 minutes is, E = P * t.
= 18.75 * 0.05 = 0.9375 Wh OR make it as round figure = 0.938 Wh.
= 225/12 = 18.75,
t = 3 minutes = (3/60) Hours = 0.05 Hours,
So, ENERGY consumption in 3 minutes is, E = P * t.
= 18.75 * 0.05 = 0.9375 Wh OR make it as round figure = 0.938 Wh.
(1)
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