Question

The current in a circuit is to be determined by measuring the voltage drop across a precision resistor in series with the circuit.

(a) What should be the resistance of the resistor in ohms if 1.00 V is to correspond to 20 μA?

(b) What must be the resistance of the voltage-measuring device if the error in the current measurement

is to be less than 2.0% relative?

Short answer

The resistance of the resistor is 50$k\Omega$and the internal resistance of the voltage measuring device is 2.45$M\Omega$

Step-by-step solution

Given information

Voltage drop is 1V and the current in the circuit is 20$\mu A$and the relative error is given to be 2%.

Part (a)

According to Ohm's Law:

V = IR

R = V/I

$$\begin{gathered} R=\frac{1}{20\times {10}^{-6}}\Omega \\ R=50\times {10}^3\Omega \\ R=50k\Omega \end{gathered}$$

Part (b)

The relative error is 2%

The expression for relative error is given by:

$E=-\frac{R}{R+R_v}\times 100$

where R_v is the resistance of the voltage measuring device.

$$\begin{gathered} -2.0=-\frac{50}{50+R_v}\times 100 \\ 50+R_v=2500k\Omega \\ {R}_v=2450k\Omega \\ {R}_v=2.45M\Omega \end{gathered}$$