Example 5.4.1.
Find the current in a wire if \(3\times 10^{20}\) electrons is passing every \(20\) second.
Section 5.4 Examples
Example 5.4.2.
Find the resistance of a light bulb that runs on a 9 V battery and draws a current of 1.5 A.
Solution.
From Ohm’s Law Subsection 5.2.2
\begin{align*}
V \amp =IR\\
or, \quad R \amp =\frac{V}{I}\\
\therefore \quad R \amp =\frac{12}{1.5} = 6 \Omega.
\end{align*}
Example 5.4.3.
If an electric iron of \(1500W\) is used for 20 minutes everyday. What is the cost of electricity bill per month, if the unit price is 15 cents?
Solution.
Here \(t\) = 20 mint/day = \(\frac{20}{60} \,h/day\times 30 \,day\) = 10 h.
Now from equation (5.2.4), we have
\begin{align*}
P \amp =\frac{E}{t}\\
or, \quad E \amp =Pt\\
or, \quad E \amp =1500\, W\times 10 \,h = 15000\,Wh = 15\,kWh
\end{align*}
In an electric bill, one unit of electrical energy is \(1\,kWh.\) Hence, the total cost of bill is \(15\,kWh\times 15 \,cnets\) = \(225 cents = $2.25\).
Example 5.4.4.
How much current is passing through the ciruit if \(1000W\) electric heater is running at 120 V?
Solution.
From equation (5.2.5)
\begin{align*}
P \amp =IV\\
or, \quad I \amp =\frac{P}{V}\\
\therefore \quad I \amp =\frac{1000}{120} = 8.33 \,A.
\end{align*}
Example 5.4.5.
What is the resistance of 50 W light bulb filament running at 120 V?
Solution.
From equation (5.2.7)
\begin{align*}
P \amp =\frac{V^2}{R}\\
or, \quad R \amp =\frac{V^2}{P}\\
or, \quad R \amp =\frac{120^2}{50}=288\Omega.
\end{align*}
Example 5.4.6.
Find how much electric energy is stored in a 10V laptop battery rated at 13000 mAh.
Hint.
Solution.
Here, \(q= 13000mAh =13000\times10^{-3}\,Ah\text{,}\) remember large unit of charge is also Ah by using eqation (5.2.1). Hence, from equation (5.4.1)
\begin{align*}
Now, \quad qV\amp =Pt\\
or, \quad Pt\amp =qV = 13000\times 10^{-3}\,Ah\times 10\,V\\
\therefore \quad E \amp = 130\,AhV
\end{align*}
by using equation (5.2.4) where, \(P=\frac{E}{t}\)
Answer.
Hence, Energy, E= 130Wh. [since, \(AV=W\)]
Example 5.4.7.
The primary coil of a transformer has 500 turns and its secondary coil has 50 turns. If the current in the secondary coil is 25 A, find the current in its primary coil.
Solution.
Given: \(N_p= 500\text{,}\) \(N_s= 50\text{,}\) \(I_s= 25 \,A\text{,}\) \(I_p= ?\)
From equation (5.3.2)
\begin{align*}
\frac{I_s}{I_p}\amp =\frac{N_p}{N_s}\\
or, I_p\amp =\frac{N_s}{N_p}I_s\\
\therefore I_p\amp =\frac{50}{500}\times 25 = 2.5\,A
\end{align*}
Example 5.4.8.
A circuit has a resistance of 20 ohms and a current of 2 amperes flowing through it. What is the voltage across the circuit?
Solution.
According to Ohm’s law, the voltage (V) across a circuit is equal to the current (I) multiplied by the resistance (R):
\begin{equation*}
V = I \times R.
\end{equation*}
Using this formula, we can calculate the voltage across the circuit as follows:
\begin{align*}
V \amp =I\times R\\
\amp = 2A\times 20\Omega\\
\amp = 40 \,Volts.
\end{align*}
Therefore, the voltage across the circuit is 40 volts.
