# Electrical Engineering - Energy and Power - Discussion

Discussion Forum : Energy and Power - General Questions (Q.No. 25)

25.

A 68 resistor is connected across the terminals of a 3 V battery. The power dissipation of the resistor is

Discussion:

6 comments Page 1 of 1.
AKASH said:
6 years ago

P=vi.

P=v(v/r),

=9/68,

=0.132W,

=132mW.

P=v(v/r),

=9/68,

=0.132W,

=132mW.

HEMRAJ Choudhar said:
9 years ago

P = v*v/r.

Then calculate the value of power dissipation.

Then calculate the value of power dissipation.

(1)

Ankit agarwal said:
10 years ago

Solution:

Here : R = 68 ohm.

V = 3 v.

P = ?.

Used formula - power (P) = VI.

But here. I = ?

Then used formula - I = V / R.

I = 3 / 68 = 0.044.

Now, p = 3*0.044 = 0.132w.

= 132 mw.

Here : R = 68 ohm.

V = 3 v.

P = ?.

Used formula - power (P) = VI.

But here. I = ?

Then used formula - I = V / R.

I = 3 / 68 = 0.044.

Now, p = 3*0.044 = 0.132w.

= 132 mw.

(1)

Akash said:
1 decade ago

Power = (V)^2/R.

Power = (3*3)/68 = 9/68 = 0.132Watts.

Or 0.132*1000 = 132 mWatts.

Power = (3*3)/68 = 9/68 = 0.132Watts.

Or 0.132*1000 = 132 mWatts.

Hari said:
1 decade ago

p=vi

And i=v/r

i.e p=v*v/r

P=3*3/68

=0.132w

=132mw

And i=v/r

i.e p=v*v/r

P=3*3/68

=0.132w

=132mw

Aniket said:
1 decade ago

What is the formula to find resistance?

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