Question

Find the net resistance of the network of resistors connected between \(A\) and \(B\), as shown in the figure.

Hint:

For complicated resistor networks:
• First identify obvious series combinations.
• Then identify parallel combinations.
• Simplify the circuit step-by-step instead of attempting the entire circuit at once.
• Always redraw the circuit after each simplification. For two resistors in parallel: \[ \boxed{ R_{\text{eq}} = \frac{R_1R_2}{R_1+R_2} } \] For two resistors in series: \[ \boxed{ R_{\text{eq}} = R_1+R_2 } \]

Answer 1

Concept: To determine the equivalent resistance of a complicated resistor network, we simplify the circuit step-by-step by identifying:
• Resistors connected in series,
• Resistors connected in parallel,
• Symmetrical combinations, if any. Two resistors \(R_1\) and \(R_2\) connected in series have an equivalent resistance \[ R_s=R_1+R_2. \] Two resistors \(R_1\) and \(R_2\) connected in parallel have an equivalent resistance \[ R_p=\frac{R_1R_2}{R_1+R_2}. \] We shall simplify the given network from left to right.

Step 1: Identify the resistors between points \(M\) and \(P\).
Between \(M\) and \(P\), there are two possible paths:
• Upper branch: one resistor of resistance \(R\).
• Lower branch: two resistors of resistance \(R\) each in series. Therefore, the resistance of the lower branch is \[ R+R=2R. \] Hence, between \(M\) and \(P\), we have two resistances \(R\) and \(2R\) connected in parallel. Their equivalent resistance is \[ R_{MP} = \frac{R(2R)}{R+2R} = \frac{2R^2}{3R} = \frac{2R}{3}. \] Thus, \[ \boxed{ R_{MP}=\frac{2R}{3} } \]

Step 2: Identify the resistors between points \(P\) and \(N\).
Between \(P\) and \(N\), there are again two branches:
• Upper branch: one resistor of resistance \(R\).
• Lower branch: one resistor of resistance \(R\). Since these two resistors are connected in parallel, \[ R_{PN} = \frac{R\times R}{R+R} = \frac{R^2}{2R} = \frac{R}{2}. \] Therefore, \[ \boxed{ R_{PN}=\frac{R}{2} } \]

Step 3: Redraw the simplified circuit.
After simplification, the circuit becomes a series combination of: \[ 2R, \qquad \frac{2R}{3}, \qquad \frac{R}{2}, \qquad 3R. \] Since all these equivalent resistances are connected in series, the total resistance between \(A\) and \(B\) is \[ R_{AB} = 2R+\frac{2R}{3}+\frac{R}{2}+3R. \]

Step 4: Add all the resistances.
Taking the LCM of \(1,3,\) and \(2\), we get \(6\). Therefore, \[ R_{AB} = \frac{12R}{6} + \frac{4R}{6} + \frac{3R}{6} + \frac{18R}{6}. \] Hence, \[ R_{AB} = \frac{37R}{6}. \] Therefore, the net resistance of the network between points \(A\) and \(B\) is \[ \boxed{ R_{AB}=\frac{37R}{6} } \]