Question

as the resistance increases, the voltage drop across the potentiometer decreases and the current flow decreases. select one: true false

Explanation:

According to Ohm's Law \(V = IR\). If the resistance \(R\) increases, assuming the voltage source (total voltage) is constant, the current \(I=\frac{V}{R}\) decreases. But for a potentiometer in a circuit (assuming it's part of a series - parallel combination), the voltage drop across it \(V_p = IR_p\). When \(R\) (total resistance in the circuit) increases, \(I\) decreases. However, if the potentiometer's resistance \(R_p\) is adjusted (say in a voltage - divider circuit), the relationship is more complex. In a simple series circuit with a fixed voltage source \(V\), total resistance \(R_{total}=R_1 + R_{pot}\). Current \(I=\frac{V}{R_{total}}\). The voltage drop across the potentiometer \(V_{pot}=IR_{pot}=\frac{VR_{pot}}{R_1 + R_{pot}}\). As \(R_{pot}\) increases, \(V_{pot}\) first increases (when \(R_{pot}

Answer:

False