Voltage divider with equal resistances limits output to half the supply

Steps:

• Total resistance in series: fixed 20Ω + potentiometer 20Ω = 40Ω.
• Current from battery: 12V / 40Ω = 0.3A.
• Potentiometer varies from 0Ω to 20Ω, so output voltage across it: I × R_pot = 0.3A × (0 to 20Ω) = 0V to 6V.
• Maximum output is half the battery voltage due to equal total resistances.

Why A is correct:

• Ohm's law and voltage divider rule: V_out = V_in × (R_pot / (R_fixed + R_pot)), max 12V × (20/40) = 6V.

Why the others are wrong:

• B: Exceeds 6V max, as fixed resistor drops half the voltage.
• C: Minimum is 0V, not 6V, when potentiometer is at 0Ω.
• D: Impossible, as battery is only 12V and no 20V source exists.

Final answer: A