Example 4.3.2. Series Resistors.
Consider the series circuit in Figure 4.3.1.
If the battery supplies 48 V, and the resistors are \(R_1 = 10\ \Omega\text{,}\) \(R_2 = 20\ \Omega\text{,}\) and\(R_3= 50\ \Omega\text{,}\) determine
- The current in the circuit, and
- The voltage drop across each resistor.
Answer.
I = 0.6 A, \(E_1\) = 6 V, \(E_2\) = 12 V, \(E_3\) = 30 V.
Solution.
The total resistance is the sum of the individual resistances
\begin{equation*}
R_t = R_1 + R_2 + R_3 = 80\ \Omega\text{.}
\end{equation*}
By Ohms Law, the current is
\begin{equation*}
I = E/R_t = \frac{48 \text{ V}}{80 \Omega} = 0.6 \text{ A}\text{.}
\end{equation*}
The voltage drop across each resistor is
\begin{align*}
E_1 \amp= I R_1= 6 \text{ V} \\
E_2 \amp= I R_2 = 12 \text{ V}\\
E_3 \amp= I R_3 = 30 \text{ V}
\end{align*}
Note that the three voltage drops add up to the 48 V supply.
