# 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?

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.

(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.

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.

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)

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:
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.

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.

(3)

Satish mahadu kukade said:
7 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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