In the circuit shown, the current in the 1Ω resistor is:

Step 1 : Applying KVL in loop 1
9 - 6i - 7(i - i₁) = 0 ⇒ 6i - 7i + 7i₁ = 9
⇒ -i + 7i₁ = 9 (1)
Step 2 : Applying KVL in loop 2
6 - 7i₁ + 1(i - i₁) = 0 ⇒ -i + 8i₁ = 6.. (2)
Step 3 : Solving equations (1) and (2)
Eq(1) × 8 + Eq(2) ⇒ -8i + 56i₁ + (-i) + 8i₁ = 72 + 6
⇒ -9i + 64i₁ = 78 (3)
Eq(1) × 7 + Eq(2) ⇒ -7i + 49i₁ - i + 8i₁ = 63 + 6
⇒ -8i + 57i₁ = 69 (4)
From (3), i = (64i₁ - 78)/9
Substituting this in (4):
-8 * (64i₁ - 78)/9 + 57i₁ = 69
-512i₁ + 624 + 513i₁ = 621
1i₁ = -3
i₁ = -3A
Put the value of i₁ in equation(1)
-i + 7(-3) = 9
-i -21 = 9
-i = 30
i = -30A
Current through 1Ω resistor = i - i₁ = -30 - (-3) = -27A
Therefore, current through 1Ω resistor is 0.13A from Q to P
Hence, Option (C) is correct.