A 1.5m wire carries a 4 A current when a potential difference of 66 V is applied. What is the resistance of the wire?
1 answer:
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Answer:
I = 2.33 µA
Explanation:
given,
Effective resistance (R)= 300 Ω
Voltage of signal (V)= 700 µV
current flow in the antenna = ?
Using ohm's law
V = I R
where I is the current flow
I = 2.33 x 10⁻⁶ A
I = 2.33 µA
Hence, the current flow in the antenna downloaded is equal to I = 2.33 µA
Answer:
The possible minimum current and maximum current through the resistor is 4.53 μA and 4.18 μA
Explanation:
Given that,
Resistance = 690 kΩ
Total voltage = 3.00 V
We need to calculate the maximum resistance
We need to calculate the minimum resistance
We need to calculate the maximum and minimum current
Using ohm's law
For maximum current,
For minimum current,
Hence, The possible minimum current and maximum current through the resistor is 4.53 μA and 4.18 μA