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NVAMediumJEE 2023Combination of Resistors

JEE Physics 2023 Question with Solution

In the given circuit, the equivalent resistance between the terminal A and B is _____ Ω\Omega.

Circuit between terminals A and B with resistors of 3 ohm and 2 ohm on top branch, 6 ohm on bottom branch, two vertical 2 ohm resistors, and two 4 ohm resistors on the right loop.

Answer

Correct answer:10

Step-by-step solution

Standard Method

Given: The circuit between terminals A and B.

Find: The equivalent resistance ReqR_{\text{eq}} between A and B.

Solution diagram showing the original circuit, note that both 4 ohm resistors are shorted, and the reduced circuit with 3 ohm, 2 ohm, and 6 ohm resistors remaining.

From the circuit reduction shown, both 4 Ω4 \, \Omega resistors are shorted, so they do not contribute to the equivalent resistance.

Now the remaining network gives

Req=3+(2∥2)+6R_{\text{eq}} = 3 + (2 \parallel 2) + 6

The parallel combination is

2∥2=2×22+2=1 Ω2 \parallel 2 = \frac{2 \times 2}{2 + 2} = 1 \, \Omega

Substituting,

Req=3+1+6=10 ΩR_{\text{eq}} = 3 + 1 + 6 = 10 \, \Omega

Therefore, the equivalent resistance between A and B is 10 Ω10 \, \Omega.

Using the observation about shorted resistors

Given: The circuit between terminals A and B.

Find: The equivalent resistance ReqR_{\text{eq}}.

Observe the circuit carefully. The two 4 Ω4 \, \Omega resistors are connected across points that are already joined by an ideal wire path, so the potential difference across each of them is zero. Hence, no current flows through them and they can be removed.

After removing those resistors, the circuit reduces to a series combination of 3 Ω3 \, \Omega, the middle parallel part (2∥2)(2 \parallel 2), and 6 Ω6 \, \Omega.

So,

Req=3+(2∥2)+6R_{\text{eq}} = 3 + (2 \parallel 2) + 6

with

2∥2=2×22+2=1 Ω2 \parallel 2 = \frac{2 \times 2}{2 + 2} = 1 \, \Omega

Thus,

Req=3+1+6=10 ΩR_{\text{eq}} = 3 + 1 + 6 = 10 \, \Omega

Therefore, the numerical value of the equivalent resistance is 10.

Common mistakes

  • Treating the two 4 Ω4 \, \Omega resistors as active elements is incorrect because they are shorted by ideal wires and have zero potential difference across them. First identify branches with no current before combining resistors.

  • Adding the two 2 Ω2 \, \Omega resistors in series is wrong. They are connected between the same two nodes in the reduced circuit, so they are in parallel. Use R=2×22+2R = \frac{2 \times 2}{2 + 2}, not 2+22 + 2.

  • Combining resistors before simplifying the short-circuited part can lead to an incorrect network. Redraw the circuit after removing branches with zero current, then apply series-parallel reduction step by step.

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