What is the potential difference across 100 Ω resistor in the circuit given below?

What is the potential difference across 100 Ω resistor in the circuit given below?

Solution:
Let,
\(R_1\) = 100 \(\Omega\)
\(R_2\) = 50 \(\Omega\)
\(R_3\) = 500 \(\Omega\)
V = 12 V \(\Omega\)

Since \(R_1\) and \(R_2\) are parallel, the equivalent resistance is, \[\begin{align*} \frac{1}{R_{12}}&=\frac{1}{R_1}+\frac{1}{R_2}\\ \frac{1}{R_{12}}&=\frac{3}{100}\\ \therefore R_{12}&=\frac{100}{3}\\ \end{align*}\] Total resistance in the circuit is then, \[\begin{align*} R&=R_{12}+R_3\\ &=\frac{100}{3}+500\\\ &=\frac{1600}{3}\\ \therefore R&=\frac{1600}{3} \hspace{0.1cm} \Omega\\ \end{align*}\] Now, \[\begin{align*} V&=IR\\ 12&=I\times \frac{1600}{3}\\ I&=\frac{36}{1600} \hspace{0.1cm} A\\ \end{align*}\] Then, voltage across 100 \(\Omega\) resistor is, \[\begin{align*} &=I\times R_{12}\\ &=\frac{36}{1600} \times \frac{100}{3}\\ &=0.75 \hspace{0.1cm} V\\ \end{align*}\]
Hence, the potential difference across 100 \(\Omega\) resistor is 0.75 V.