A battery of 9V is connected in series with resistors of 0.2 ohms, 0.3 ohms, 0.4 ohms, 0.5 ohms, and 12 ohms respectively. How much current would flow through the 12 ohms resistor?

Detailed Solution

To solve this question, let's break it down into simple steps.

Step 1: Understand the question

We are given:

• A battery of 9V is connected in series with several resistors: 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω, and 12 Ω.
• We need to calculate the current flowing through the 12 Ω resistor.

Step 2: Key concept

When resistors are connected in series, the same current flows through all the resistors. So, the current flowing through the 12 Ω resistor is the same as the current flowing through the entire circuit.

Step 3: Find the total resistance

In a series connection, the total resistance (Rtotal) is the sum of all the resistances:

Rtotal = R1 + R2 + R3 + R4 + R5

Substitute the given resistances:

Rtotal = 0.2 + 0.3 + 0.4 + 0.5 + 12

Calculate:

Rtotal = 13.4 Ω

Step 4: Use Ohm's Law

Ohm's Law states:

I = V / R

Where:

• I is the current
• V is the voltage of the battery
• R is the total resistance

Here, V = 9 V and R = 13.4 Ω.

Substitute the values:

I = 9 / 13.4

Calculate:

I = 0.67 A

Step 5: Conclusion

Since the resistors are connected in series, the current through each resistor, including the 12 Ω resistor, is the same. Therefore, the current through the 12 Ω resistor is:

I = 0.67 A

Final Answer: 0.67 A

Key points to remember:

1. A battery of 9V is connected in series with all the resistors, meaning the same current flows through all of them.
2. In a series circuit, the total resistance is the sum of all individual resistances.
3. Using Ohm's Law, we calculated the current to be 0.67 A.

Thus, the correct answer is (a) 0.67 A.