Question

The current in $7.2 \, \Omega$ resistor is :

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

Hint:

In series circuits, the current is the same through all components. In parallel circuits, the voltage across each branch is the same. Always simplify parallel combinations first, then add series resistances to find total resistance.

Answer 1

The Correct Option is C

Step 1: Identify the components and configuration in the circuit.
The circuit consists of a 12V battery, a $7.2 \, \Omega$ resistor in series with a parallel combination of an $8 \, \Omega$ resistor and a $12 \, \Omega$ resistor. The current in the $7.2 \, \Omega$ resistor is the total current flowing out of the battery, as it is in the main series path. Step 2: Calculate the equivalent resistance of the parallel combination.
For two resistors in parallel, the equivalent resistance ($R_{eq,parallel}$) is given by:
$\frac{1}{R_{eq,parallel}} = \frac{1}{R_1} + \frac{1}{R_2}$
Here, $R_1 = 8 \, \Omega$ and $R_2 = 12 \, \Omega$.
$\frac{1}{R_{eq,parallel}} = \frac{1}{8} + \frac{1}{12} = \frac{3 + 2}{24} = \frac{5}{24}$
$R_{eq,parallel} = \frac{24}{5} = 4.8 \, \Omega$ Step 3: Calculate the total equivalent resistance of the circuit.
The $7.2 \, \Omega$ resistor is in series with the parallel combination ($4.8 \, \Omega$).
For resistors in series, the total equivalent resistance ($R_{total}$) is the sum of individual resistances:
$R_{total} = R_{series} + R_{eq,parallel}$
$R_{total} = 7.2 \, \Omega + 4.8 \, \Omega = 12 \, \Omega$
Step 4: Calculate the total current flowing through the circuit (and thus through the $7.2 \, \Omega$ resistor).
Using Ohm's Law, $V = IR$, where $V$ is the total voltage, $I$ is the total current, and $R$ is the total equivalent resistance.
$I = \frac{V}{R_{total}}$
Given $V = 12 \, \text{V}$ and $R_{total} = 12 \, \Omega$.
$I = \frac{12 \, \text{V}}{12 \, \Omega} = 1 \, \text{A}$
Step 5: Conclude the current in the $7.2 \, \Omega$ resistor.
Since the $7.2 \, \Omega$ resistor is in the main series path, the current flowing through it is the total current of the circuit, which is 1 A. (3) 1 A