Question

What is the maximum current that flows in the thread during this experiment if the voltage source is a 9-V battery? (a) about 1 A; (b) about 0.1 A; (c) about 1$\mathit{\mu }$A; (d) about 1 nA.

Short answer

Option (d) is correct.

Step-by-step solution

Ohm’s law

Electric current is proportional to voltage and inversely proportional to resistance, according to Ohm's law.

$I_{max}=\frac{V}{R}$

Solution

For the maximum current;

Least Resistance$9\times {10}^9\Omega$

Voltage$V=9.0V$

Putting the values;

$$\begin{gathered} I_{max}=\frac{V}{R} \\ {I}_{max}=\frac{9.0V}{9\times {10}^9\Omega } \\ =1.0\times {10}^{-9}A=1.0nA \end{gathered}$$

Hence, option (d) is correct.