Question 22 of 31intermediate🔧 ApplyNumerical3 marks

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

Correct Answer

Step 1: Calculate the total resistance. Since the resistors are connected in series, the total resistance (RsR_s) is equal to the sum of the individual resistances.

Rs=0.2 Ω+0.3 Ω+0.4 Ω+0.5 Ω+12 ΩR_s = 0.2 \ \Omega + 0.3 \ \Omega + 0.4 \ \Omega + 0.5 \ \Omega + 12 \ \Omega

Rs=13.4 ΩR_s = 13.4 \ \Omega.

Step 2: Calculate the total current in the circuit. Using Ohm's law, the current II is given by the formula I=V/RI = V / R. Given potential difference, V=9 VV = 9 \text{ V}. I=9 V/13.4 Ω=0.67 AI = 9 \text{ V} / 13.4 \ \Omega = 0.67 \text{ A} (approx).

Step 3: Determine current through the 12 Ω\Omega resistor. In a series circuit, the same current flows through every resistor. Therefore, the current flowing through the 12 Ω12 \ \Omega resistor is 0.67 A.

Exercise: EXERCISES | Q: 9 | (Page Page 23)
For More Understanding

Explanation

The context establishes that for a series circuit, the total resistance is the sum of individual resistances (as illustrated in the example where resistances were added). It also provides the basis for applying Ohm's Law (I=V/RI = V/R) to find the current. Because the components are in series, there is only one path for the current to flow, meaning the calculated total current is the same current that passes through the 12 Ω resistor.

Solution Steps

  1. Step 1: Sum the individual resistances to find total resistance (Rs=13.4 ΩR_s = 13.4 \ \Omega).

  2. Step 2: Apply Ohm's Law (I=V/RI = V/R) using the battery voltage and total resistance to find total current.

  3. Step 3: Conclude that in a series connection, the current through the specific resistor is equal to the total current.

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