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
resistor is connected across the terminals of a 3 V battery. The power dissipation of the resistor isDiscussion:
6 comments Page 1 of 1.
AKASH said:
7 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:
10 years ago
P = v*v/r.
Then calculate the value of power dissipation.
Then calculate the value of power dissipation.
(1)
Ankit agarwal said:
1 decade 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.
(2)
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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