Six equal resistances are connected between points P, Q and R as shown in the figure. Then, the net

Q: Why are there multiple zig-zag lines on some sides?

Each zig-zag line represents a separate resistor. Multiple lines drawn between the same two nodes indicate that those resistors are connected in parallel.

Q: How do we know the resistors are in parallel and not in series?

The lines start at one common node and end at another common node without any other connections in between. This is the exact definition of a parallel connection.

Q: Can we use star-delta transformation here?

While you could use star-delta if the circuit was more complex, here it's unnecessary. Once you simplify the parallel branches on each side, you are left with a simple triangular loop where any two nodes form a parallel circuit of one side and the sum of the other two.

Q: Why is the resistance maximum across the branch with the highest individual resistance?

In a parallel circuit, the equivalent resistance is heavily influenced by the smallest resistance. However, in this specific triangular topology, the branch with the highest direct resistance (PQ) forces the current to take the alternate path, which also has a relatively high resistance, resulting in the highest overall equivalent resistance.

Q: What if all sides had the same number of resistors?

If all sides had the same number of resistors, the circuit would be perfectly symmetric. In that case, the equivalent resistance between any two nodes would be identical.

Question

Six equal resistances are connected between points $P, Q$ and $R$ as shown in the figure. Then, the net resistance will be maximum between

Options:

(A) $P$ and $Q$

(B) $Q$ and $R$

(C) $P$ and $R$

(D) any two points

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When faced with a seemingly complex circuit diagram, the first step is always to decode the visual information into a standard electrical topology. In this problem, we are given a triangular network with nodes $P$, $Q$, and $R$.

If you look closely at the zig-zag lines representing the resistors, you will notice they are not uniformly distributed. 
- Between nodes $P$ and $Q$, there is exactly **one** resistor.
- Between nodes $Q$ and $R$, there are **two** resistors drawn side-by-side, indicating they are in parallel.
- Between nodes $P$ and $R$, there are **three** resistors drawn side-by-side, also in parallel.

Let's assume each individual resistor has a resistance of $r$. We can simplify each side of the triangle by finding the equivalent resistance of these parallel groups.

### Simplifying the Branches

For the branch between $P$ and $Q$, the equivalent resistance is simply:
$$R_1 = r$$

For the branch between $Q$ and $R$, we have two resistors in parallel. Their equivalent resistance is:
$$R_2 = \frac{r}{2}$$

For the branch between $P$ and $R$, we have three resistors in parallel. Their equivalent resistance is:
$$R_3 = \frac{r}{3}$$

Now, our complex network has been reduced to a simple triangular loop with three equivalent resistors: $R_1$, $R_2$, and $R_3$.

### Calculating Equivalent Resistances

To find the net resistance between any two nodes in a triangular loop, we must recognize that the direct branch connecting the two nodes is in parallel with the series combination of the other two branches.

**1. Resistance between $P$ and $Q$:**
The direct path is $R_1$. The alternate path goes through node $R$, meaning $R_2$ and $R_3$ are in series.
$$R_{PQ} = R_1 \parallel (R_2 + R_3)$$
$$R_{PQ} = r \parallel \left(\frac{r}{2} + \frac{r}{3}\right) = r \parallel \frac{5r}{6}$$
Using the product-over-sum rule:
$$R_{PQ} = \frac{r \cdot \frac{5r}{6}}{r + \frac{5r}{6}} = \frac{\frac{5r}{6}}{\frac{11}{6}} = \frac{5}{11}r$$

**2. Resistance between $Q$ and $R$:**
The direct path is $R_2$, which is in parallel with the series combination of $R_1$ and $R_3$.
$$R_{QR} = R_2 \parallel (R_1 + R_3)$$
$$R_{QR} = \frac{r}{2} \parallel \left(r + \frac{r}{3}\right) = \frac{r}{2} \parallel \frac{4r}{3}$$
$$R_{QR} = \frac{\frac{r}{2} \cdot \frac{4r}{3}}{\frac{r}{2} + \frac{4r}{3}} = \frac{\frac{4r}{6}}{\frac{11r}{6}} = \frac{4}{11}r$$

**3. Resistance between $P$ and $R$:**
The direct path is $R_3$, which is in parallel with the series combination of $R_1$ and $R_2$.
$$R_{PR} = R_3 \parallel (R_1 + R_2)$$
$$R_{PR} = \frac{r}{3} \parallel \left(r + \frac{r}{2}\right) = \frac{r}{3} \parallel \frac{3r}{2}$$
$$R_{PR} = \frac{\frac{r}{3} \cdot \frac{3r}{2}}{\frac{r}{3} + \frac{3r}{2}} = \frac{\frac{r}{2}}{\frac{11r}{6}} = \frac{3}{11}r$$

### Conclusion

Comparing the three equivalent resistances, we have:
$$R_{PQ} = \frac{5}{11}r$$
$$R_{QR} = \frac{4}{11}r$$
$$R_{PR} = \frac{3}{11}r$$

Since $\frac{5}{11} > \frac{4}{11} > \frac{3}{11}$, the net resistance is **maximum between points $P$ and $Q$**.

Animated Solution for Physics - Current Electricity: Six equal resistances are connected between points $P, Q$ and $R$ as shown in the figure. Then, the net resistance will be maximum between

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Visualized Solution

$Analyzing the Circuit Topology$

$Equivalent Resistance of Each Branch$

$Net Resistance Between P and Q$

$Calculating R_{P Q}$

$Net Resistance Between Q and R$

$Calculating R_{QR}$

$Net Resistance Between P and R$

$Calculating R_{P R}$

$Comparing the Resistances$

$Conclusion$

The Sigma Insight: Combination of Resistors

Analyzing the Setup

Simplifying the Branches

Calculating Equivalent Resistances

Conclusion

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Circuit Topology

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equivalent Resistance of Each Branch

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Resistance Between P and Q

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating RPQ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Resistance Between Q and R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating RQR​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Resistance Between P and R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating RPR​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Comparing the Resistances

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion