Question

A charge of $4.00C$ charge passes through a pocket calculator’s solar cells in $4.00h$. What is the power output, given the calculator’s voltage output is $3.00V$? (See Figure $20.40$.)

Short answer

The power output of the calculator is obtained as: $P=0.833mW$.

Step-by-step solution

Define Resistance

In an electrical circuit, resistance is a measure of the resistance to current flow. The Greek letter omega $(\mathrm{\Omega })$ is used to represent resistance in ohms.

Concepts and Principles

Electric current is the magnitude of the physical quantity of electric current $\mathrm{I}$ in a wire which equals the magnitude of the electric charge $q$ that passes through a cross-section of the wire divided by the time interval $\mathrm{\Delta }t$ needed for that charge to pass:

$I=\frac{|q|}{\mathrm{\Delta }t}$……………..(I)

The unit of current is the ampere$A$, which is equivalent to one coulomb per second$(C/s)$.A current of$1A$(one ampere, or amp) means that$1C$of charge passes through a cross-section of the wire every second. The direction of the current is from higher to lower potential. So, it is in the direction that positive charges that would move.

If a potential difference $\mathrm{\Delta }V$ is maintained across a circuit element, the power, or the rate at which energy is supplied to the element, is:$P=I\mathrm{\Delta }V$………….(II)

The given data and the required data

• Charge passing through the calculator is: $q=4.00C$.
• The time interval needed for that charge to pass is:$\begin{array}{rl}\mathrm{\Delta }t& =(4.00h)(\frac{3600s}{1h})\\ & =14400s\end{array}$

• The voltage output of the calculator is: $\mathrm{\Delta }V=3.00V$.

Evaluating the power output of the calculator

Current in the calculator is found by using Ohm's law with the help of the first equation as:

$I=\frac{q}{\mathrm{\Delta }t}$

Entering values and we obtain:

$\begin{array}{rl}I& =\frac{4.00C}{14400s}\\ & =\frac{1}{3600}A\end{array}$

The power output of the calculator is found with the help of the second equation as:

$P=I\mathrm{\Delta }V$

Substitute numerical values, and we get:

$\begin{array}{rl}P& =(\frac{1}{3600}A)(3.00V)\\ & =8.33\times {10}^{-4}W\\ & =(8.33\times {10}^{-4}W)(\frac{1000mW}{1W})\\ & =0.833mW\end{array}$

Therefore, the power output is: $P=0.833mW$.