Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance \( R_p = 1 \, \Omega \) as shown in the figure. An external resistance of \( R_e = 2 \, \Omega \) is connected via the sliding contact. The current \( i \) is :
Show Hint
For resistors in series and parallel, simplify the equivalent resistance and then use Ohm's law to calculate the current.
The equivalent resistance of the circuit is calculated as: \[ R_{\text{eq}} = 0.5 + \frac{0.5 \times 2}{2 + 0.5} = \frac{5}{10} + \frac{10}{25} \, \Omega = 0.9 \, \Omega \] The current flowing through the circuit is then determined by: \[ i = \frac{0.9}{0.9} = 1 \, \text{A} \] Consequently, the current is \( \boxed{1.0 \, \text{A}} \).
Was this answer helpful?
2
Top Questions on Semiconductor electronics: materials, devices and simple circuits
