Wheatstone Bridge Balance Condition Steps:

• The potentiometer divides into lower resistance 50Ω (1/4 of 200Ω) and upper resistance 150Ω.
• In the bridge, one ratio arm is upper/lower = 150Ω/50Ω = 3.
• The other ratio arm has known resistor 150Ω and unknown R.
• At null deflection (0V), 150/50 = 150/R, so R = 150/3 = 50Ω.

Why C is correct:

• Balance condition P/Q = S/R gives R = (Q × S)/P = (50 × 150)/150 = 50Ω, per Wheatstone's law.

Why the others are wrong:

• A: 30Ω would balance at lower resistance 100Ω (150/100 = 150/30? No, 1.5 ≠ 5).
• B: 38Ω mismatches ratio (150/50 = 3, but 150/38 ≈ 3.95 ≠ 3).
• D: 450Ω inverts ratio (150/450 = 1/3, but needs 3).

Final answer: C