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Given an ohmmeter with a 50 μA, 10000 Ω meter gauge and a 3 V DC supply, what is the ohmmeter's total resistance? Solve for the resistance in kΩ to 2 decimal places of accuracy.

Answer :

Final answer:

The total resistance of the ohmmeter is 60 kiloohms (ko) to 2 decimal places of accuracy.

Explanation:

An ohmmeter is a device used to measure the electrical resistance of a circuit component. It consists of a meter gauge and an internal resistance. When measuring the resistance of a component, the ohmmeter is connected in parallel with the component. The meter gauge measures the current flowing through the component, and the internal resistance of the ohmmeter is adjusted to ensure accurate readings.

To calculate the total resistance of the ohmmeter, we can use Ohm's Law, which states that the resistance (R) is equal to the voltage (V) divided by the current (I).

Given:

  • Current (I) = 50uA = 50 * 10^-6 A
  • Voltage (V) = 3V

Using Ohm's Law:

R = V / I

R = 3V / (50 * 10^-6 A)

R = 3V / 0.00005 A

R = 60,000 ohms

To convert the resistance to kiloohms (ko), we divide by 1000:

R = 60,000 ohms / 1000 = 60 ko

Therefore, the total resistance of the ohmmeter is 60 kiloohms (ko) to 2 decimal places of accuracy.

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