Question

Calculate total current flowing through the circuit if \(R_1 = 7.2 \, \Omega\), \(R_2 = 8 \, \Omega\) and \(R_3 = 12 \, \Omega\) :

• A. 2 A
• B. 1.5 A
• C. 1 A
• D. 0.5 A

Hint:

1. Identify series and parallel combinations. Here \(R_2\) and \(R_3\) are in parallel, and their combination is in series with \(R_1\). 2. Calculate equivalent resistance for parallel \(R_2\) and \(R_3\): \(R_P = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{8 \times 12}{8 + 12} = \frac{96}{20} = \frac{24}{5} = 4.8 \, \Omega\). (Alternatively, \(\frac{1}{R_P} = \frac{1}{8} + \frac{1}{12} = \frac{3+2}{24} = \frac{5}{24} \Rightarrow R_P = \frac{24}{5} = 4.8 \, \Omega\)). 3. Calculate total series resistance: \(R_{eq} = R_1 + R_P = 7.2 + 4.8 = 12 \, \Omega\). 4. Use Ohm's Law (\(I = V/R\)) for total current: \(I = \frac{6 \text{ V}}{12 \, \Omega} = 0.5 \text{ A}\).

Answer 1

The Correct Option is D

Concept: To find the total current, we first need to find the total equivalent resistance (\(R_{eq}\)) of the circuit. Then, using Ohm's Law (\(V = IR\)), we can find the total current \(I = V/R_{eq}\). The diagram shows \(R_2\) and \(R_3\) are connected in parallel, and this parallel combination is connected in series with \(R_1\). The voltage of the source is 6V. Step 1: Calculate the equivalent resistance of the parallel combination of \(R_2\) and \(R_3\) Let \(R_P\) be the equivalent resistance of \(R_2\) and \(R_3\) in parallel. The formula for two resistors in parallel is \(\frac{1}{R_P} = \frac{1}{R_2} + \frac{1}{R_3}\). Given \(R_2 = 8 \, \Omega\) and \(R_3 = 12 \, \Omega\). \[ \frac{1}{R_P} = \frac{1}{8} + \frac{1}{12} \] To add these fractions, find a common denominator, which is 24. \[ \frac{1}{R_P} = \frac{3}{24} + \frac{2}{24} = \frac{3+2}{24} = \frac{5}{24} \] So, \(R_P = \frac{24}{5} \, \Omega\). \[ R_P = 4.8 \, \Omega \] Step 2: Calculate the total equivalent resistance (\(R_{eq}\)) of the circuit The resistor \(R_1\) is in series with the parallel combination \(R_P\). For resistors in series, the total resistance is the sum of individual resistances. \[ R_{eq} = R_1 + R_P \] Given \(R_1 = 7.2 \, \Omega\) and we found \(R_P = 4.8 \, \Omega\). \[ R_{eq} = 7.2 \, \Omega + 4.8 \, \Omega \] \[ R_{eq} = 12.0 \, \Omega \] Step 3: Calculate the total current (\(I\)) flowing through the circuit using Ohm's Law Given voltage of the source, \(V = 6 \text{ V}\). Total equivalent resistance, \(R_{eq} = 12 \, \Omega\). Using Ohm's Law, \(I = \frac{V}{R_{eq}}\): \[ I = \frac{6 \text{ V}}{12 \, \Omega} \] \[ I = \frac{1}{2} \text{ A} \] \[ I = 0.5 \text{ A} \] The total current flowing through the circuit is 0.5 A. This matches option (4).