# 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?
938 Wh
0.938 Wh
56.25 Wh
5.6 Wh
Explanation:
No answer description is available. Let's discuss.
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.
(2)

Dhina said:   3 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.
(10)

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.
(3)

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.
(4)

Bibek said:   6 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.

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.
(3)

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.
(1)