Electrical Engineering - Energy and Power - Discussion

Discussion Forum : Energy and Power - General Questions (Q.No. 24)
24.
When the current through a 12 k resistor is 8 mA, the power is
7.68 mW
768 mW
7.68 W
76.8 W
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
3 comments Page 1 of 1.

DINESH said:   2 months ago
W = I^2R.
W = 8^2 * 12000,
W = 768000/1000,
W = 768mW.

Ashim BHowmick said:   9 years ago
V = IR = 8 mA x 12 k.ohms = 0.008 A x 12000 ohm = 96 V.
P = VI = 96 x 0.008 = 0.768 W = 768 mW.

Ketan said:   1 decade ago
P= I2R watt
P= (0.008)2 x (12*1000) watt
P= 768 x 10^-3 watt
P= 768 mW

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