Mastering Circuits!
We will solve some questions of Series and Parallel Circuit.
Basic Formula
Figure 1 – n resistors in series
▪️Equivalent Resistance,
\small \begin{aligned} &R_{eq} = R_1 + R_2 + R_3 + \ \text{\text{-} \text{-} \text{-} \text{-}} \ + R_N \end{aligned}
Figure 2 – n inductors in series
▪️Equivalent Inductance,
\small \begin{aligned} &L_{eq} = L_1 + L_2 + L_3 + \ \text{\text{-} \text{-} \text{-} \text{-}} \ + L_N \end{aligned}
Figure 3 – n conductors in series
▪️Equivalent Conductance,
\small \begin{aligned} &\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \ \text{\text{-} \text{-} \text{-} \text{-}} \ + \frac{1}{C_N} \end{aligned}
Figure 4 – n resistors in parallel
▪️Equivalent Resistance,
\small \begin{aligned} &\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ \text{\text{-} \text{-} \text{-} \text{-}} \ + \frac{1}{R_N} \end{aligned}
Figure 5 – n inductors in parallel
▪️Equivalent Inductance,
\small \begin{aligned} &\frac{1}{L_{eq}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3} + \ \text{\text{-} \text{-} \text{-} \text{-}} \ + \frac{1}{L_N} \end{aligned}
Figure 6 – n conductors in parallel
▪️Equivalent Conductance,
\small \begin{aligned} &C_{eq} = C_1 + C_2 + C_3 + \ \text{\text{-} \text{-} \text{-} \text{-}} \ + C_N \end{aligned}
Questions Solution
Some solved questions of Series and Parallel Circuit.
Problem 1
Pb-1: Find equivalent resistance between a-b in the following circuit.
Figure 7 – Circuit diagram for Pb-1
Solution:
Simplifying the circuit,
Figure 8 – Circuit diagram for solving the Pb-1
\small \begin{aligned} &R_{ab} = (((((2 + 10)||24) + 2)||((12||12) + 4)) + 7 \\[2ex] & \scriptsize{\Rightarrow R_{ab} = \left(\left(\left((2 + 10)^{-1} + 24^{-1}\right)^{-1} + 2\right)^{-1} + \left((12^{-1} + 12^{-1})^{-1} + 4\right)^{-1}\right)^{-1} + 7} \\[2ex] &\therefore R_{ab} = 12 \ \Omega \end{aligned}
Problem 2
Pb-2: Determine R_{eq} and I in the circuit shown below.
Figure 9 – Circuit diagram for Pb-2
Solution:
\small \begin{aligned} &R_{eq} = (((20||80) + (6||12))||60||15) + 5 \\[2ex] & \scriptsize{\Rightarrow R_{eq} = \left(\left((20^{-1} + 80^{-1})^{-1} + (6^{-1} + 12^{-1})^{-1}\right)^{-1} + 60^{-1} + 15^{-1}\right)^{-1} + 5} \\[2ex] &\therefore R_{eq} = 12.5 \ \Omega \\[2ex] &\therefore I = \frac{40}{R_{eq}} = \frac{40}{12.5} = 3.2 \ A \end{aligned}
Problem 3
Pb-3: Determine R_{eq} in the following circuit.
Figure 10 – Circuit diagram for Pb-3
Solution:
Figure 11 – Circuit diagram for solving the Pb-3
The resistors connected between the same nodes, b-b, will be short circuited.
Figure 12 – Circuit diagram for solving the Pb-3 (simplifying the circuit in Fig. 11)
\small \begin{aligned} &R_{eq} = (3||6||1) + 10 \\[2ex] &\Rightarrow R_{eq} = (3^{-1} + 6^{-1} + 1^{-1})^{-1} + 10 \\[2ex] &\therefore R_{eq} = 10.67 \ \Omega \end{aligned}