Three resistance coils of 1 Q, 2 Q and 3 Q are connected in series. If the combination is connected to a battery of 9 V, what is the potential drop across the resistance coil of 3 fi ?

The correct answer is Option 3 (4.5 volt). This result is derived using Ohm's Law and the principles of series circuits.

In a series connection, the equivalent resistance (Req) is the sum of individual resistances. Here, Req = 1 Ω + 2 Ω + 3 Ω = 6 Ω. Since the components are in series, the same current (I) flows through each resistor.

Using Ohm’s Law (V = IR), the total current in the circuit is:
I = V / Req = 9 V / 6 Ω = 1.5 A.

The potential drop (V₃) across the specific 3 Ω coil is calculated by multiplying the common current by its resistance:
V₃ = I × R₃ = 1.5 A × 3 Ω = 4.5 V.

• Option 1 & 2: These represent the potential drops across the 1 Ω (1.5 V) and 2 Ω (3.0 V) coils, respectively.
• Option 4: This value is incorrect as it exceeds the proportional share of the 9V supply for a 3 Ω resistor in this specific configuration.