# A simple circuit is constructed so that a 12.0-V source supplies current to a 1.25 \times 10^3...

## Question:

A simple circuit is constructed so that a 12.0-V source supplies current to a {eq}1.25 \times 10^3 \Omega {/eq} resistor.

(a) Find the current in the resistor. A

(b) Calculate the power delivered to the resistor. W

(c) Determine the energy delivered to the resistor in 1.50 h.

## Power:

Power

The work done on the object in the unit time is known as the power generated. The power is generally expressed by the unit Watt. The mathematical expression for the power in the electrical science is {eq}P = {I^2}R {/eq} where I is current.

Given data

• The value of the voltage is {eq}V = 12\;{\rm{V}} {/eq}
• The value of the resistance is {eq}R = 1.25 \times {10^3}\;\Omega {/eq}
• The value of the time is {eq}t = 1.5\;{\rm{hr = 1}}{\rm{.5}} \times {\rm{3600}}\;{\rm{sec}} {/eq}

(a)

The expression for the current is

$$\color{red}{I = \dfrac{V}{R}}$$

Substitute the value in the above equation

\begin{align*} I &= \dfrac{{12}}{{1.25 \times {{10}^3}}}\\ &= 9.6 \times {10^{ - 3}}\;{\rm{amp}} \end{align*}

Thus the value of the current is {eq}\boxed{\color{blue}{9.6 \times {10^{ - 3}}\;{\rm{amp}}}} {/eq}

(b)

The expression for the power is

$$\color{red}{P = {I^2}R}$$

Substitute the value in the above equation

\begin{align*} P &= {\left( {9.6 \times {{10}^{ - 3}}} \right)^2} \times 1.25 \times {10^3}\\ &= 0.1152\;{\rm{W}} \end{align*}

Thus the value of the power is {eq}\boxed{\color{blue}{0.1152\;{\rm{W}}}} {/eq}

(c)

The expression for the energy delivered is

$$\color{red}{U = Pt}$$

Substitute the value in the above equation

\begin{align*} U &= 0.1152 \times 1.5 \times 3600\\ &= 622.08\;{\rm{J}} \end{align*}

Thus the value of the energy is $$\boxed{\color{blue}{622.08\;{\rm{J}}}}$$ 