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
resistor is 8 mA, the power isDiscussion:
4 comments Page 1 of 1.
SUNIL said:
3 months ago
P = I^2 * R,
P = (8 * 10^-3 * 8*10^-3) * 12 * 10^3,
P = (64 * 10^-6) *12 * 10^3,
P = 768 * 10^-3,
Therefore,
P = 768 mW.
P = (8 * 10^-3 * 8*10^-3) * 12 * 10^3,
P = (64 * 10^-6) *12 * 10^3,
P = 768 * 10^-3,
Therefore,
P = 768 mW.
Ashim BHowmick said:
1 decade 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.
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
P= (0.008)2 x (12*1000) watt
P= 768 x 10^-3 watt
P= 768 mW
DINESH said:
1 year ago
W = I^2R.
W = 8^2 * 12000,
W = 768000/1000,
W = 768mW.
W = 8^2 * 12000,
W = 768000/1000,
W = 768mW.
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