Subelement A: — Topic :
Question 6A67
Element 6 (Radiotelegraph)What is the maximum current carrying capacity of a resistor of 5000 ohms, 200 watts?
Explanation
To determine the maximum current a resistor can safely carry, we use the relationship between power, current, and resistance, derived from Ohm's Law. The relevant formula is P = I²R, where P is power in watts, I is current in amperes, and R is resistance in ohms.
We need to solve for current (I), so we rearrange the formula:
I² = P / R
I = √(P / R)
Now, plug in the given values:
P = 200 watts
R = 5000 ohms
I = √(200 W / 5000 Ω)
I = √(2 / 50)
I = √(1 / 25)
I = 1 / 5
I = 0.2 A
Therefore, the maximum current carrying capacity is 0.2 amperes. Options B and C are incorrect as they result from misapplying the formula or calculation errors.
Related Questions
6A65 The total power dissipation capability of two 10 watt, 500 ohm resistors connected in series is:6A66 What is the total power dissipation capability of two 10 watt 500 ohm resistors connected in parallel?6A68 What is the total resistance of a parallel circuit consisting of a 10 ohm branch and a 25 ohm branch?6A69 The current through two resistors in series is 36A7 Modern reserve transmitters are solid-state designs and transmit using only A2 modulation. Whenmeasuring transmitter center frequency, what precaution must be taken: