Chapter 6, Problem 9STP

To determine

To Choose: The correct option.

Answer to Problem 9STP

Option B

Explanation of Solution

Given data:

Consider the circuit diagram given in the question.

The voltage across the circuit, $V=12\text{ V}$

The resistance of the bulb A, $R_{\text{A}}=8\Omega$

The resistance of the bulb B, $R_B=15\Omega$

Formula used:

• Ohm’s law equation:

  $I=\frac{V}{R}$

Where, $I$ is the current in a circuit (in amperes), $V$ is the voltage difference across the circuit in volts and $R$ is the resistance of the circuit in ohms.

• Electrical Power Equation:

  $P=IV$

Where, $P$ is the power of a device in watts, $I$ is the current through the device in amperes, and $V$ is the voltage difference across the device in volts.

Calculation:

As per the problem,

Consider the circuit diagram given in the question.

The given circuit is a parallel circuit as it has two branches through which current flows.

In a parallel circuit, the voltage difference is the same in each branch.

Hence, voltage difference across the light bulb A, $V=12\text{ V}$

Resistance of the bulb A, $R_{\text{A}}=8\Omega$

Substitute the values of voltage difference and resistance in the ohm’s law equation:

  $\begin{array}{l}{I}_{\text{A}}=\frac{V}{R_{\text{A}}}\\ I_{\text{A}}=\frac{12\text{ V}}{\text{8 }\Omega }\\ I_{\text{A}}=1.5\text{ A}\end{array}$

To find the power of the light bulb A, substitute the values of voltage difference and current (calculated above) in the power equation:

  $\begin{array}{l}{P}_{\text{A}}=I_{\text{A}}V\\ P_{\text{A}}=(1.5\text{ A)(12 V)}\\ P_{\text{A}}=18\text{ W}\end{array}$

Conclusion:

The power of the lightbulb A in the given circuit is $\underset{\_}{18\text{ W}}$ .