In the given figure, the emf of the cell is 2.2 V and if internal resistance is 0.6 ohm. Calculate t

Q: Why are the diagonal resistors in series?

Because the crossing point in the middle is not a junction (there is no dot). The current flows straight through the diagonal wire without splitting.

Q: What if the question asked for power in the external circuit?

Then we would use the formula $P = I^2 R_{eq}$, where $I = \frac{E}{R_{eq} + r}$. The power would only be the heat dissipated by the external resistors.

Q: How did you find the LCM so quickly?

For numbers like 4, 6, 8, and 12, you can look at the largest number (12) and check its multiples. 12 is not divisible by 8. The next multiple is 24, which is divisible by 4, 6, 8, and 12.

Q: Is this a Wheatstone bridge?

It looks like one, but it's drawn differently. In a standard Wheatstone bridge, the central element connects the two parallel branches. Here, the diagonals cross without connecting, creating four independent parallel branches between the two main nodes.

Question

In the given figure, the emf of the cell is $2.2	ext{ V}$ and if internal resistance is $0.6\Omega$. Calculate the power dissipated in the whole circuit

Options:

(A) $1.32	ext{ W}$

(B) $0.65	ext{ W}$

(C) $2.2	ext{ W}$

(D) $4.4	ext{ W}$

Sigma Percentile · Accepted Answer

Imagine you are an electron standing at node A, looking across the circuit to reach node B. At first glance, the diagram looks like a tangled web, a complex bridge that might require Kirchhoff's laws to solve. But as a master of physics, you know that appearances can be deceiving. Let's take a breath and trace the paths.

### Decoding the Circuit Diagram

When we look closely at the connections, a beautiful symmetry emerges. From the left vertical wire (let's call it Node A), there are four distinct paths that lead directly to the right vertical wire (Node B).

There is the top horizontal wire, the bottom horizontal wire, and two diagonal wires that cross each other in the middle. Crucially, the crossing point of the diagonals has no junction dot! This means the wires simply pass over each other without connecting.

### The Power of Parallelism

Because all four of these paths start at Node A and end at Node B, they are perfectly in parallel. This realization transforms a seemingly impossible problem into a straightforward calculation. Let's find the resistance of each individual branch.

The top branch is simply $4\Omega$.

The first diagonal branch consists of a $4\Omega$ and an $8\Omega$ resistor in series, giving us $12\Omega$.

The second diagonal branch has a $2\Omega$ and a $4\Omega$ resistor in series, totaling $6\Omega$.

Finally, the bottom branch is an $8\Omega$ resistor.

### Calculating the Equivalent Resistance

Now, we unleash the parallel resistance formula:

$$ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} $$

Substituting our branch resistances:

$$ \frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{12} + \frac{1}{6} + \frac{1}{8} $$

To add these fractions, we find the least common multiple, which is $24$.

$$ \frac{1}{R_{eq}} = \frac{6}{24} + \frac{2}{24} + \frac{4}{24} + \frac{3}{24} = \frac{15}{24} $$

Flipping the fraction gives us the equivalent external resistance:

$$ R_{eq} = \frac{24}{15} = 1.6\Omega $$

### The Whole Circuit Power

Here is where many students fall into a trap. The question specifically asks for the power dissipated in the **whole circuit**. This means we cannot ignore the battery itself! The battery has an internal resistance $r = 0.6\Omega$, which also dissipates heat.

The total resistance of the entire circuit is the sum of the external equivalent resistance and the internal resistance:

$$ R_{total} = R_{eq} + r = 1.6\Omega + 0.6\Omega = 2.2\Omega $$

The total power dissipated by the entire system is given by the square of the electromotive force (EMF) divided by the total resistance:

$$ P_{total} = \frac{E^2}{R_{total}} $$

### The Final Strike

We are now ready for the final execution. We substitute our known values into the master equation:

$$ P_{total} = \frac{(2.2)^2}{2.2} $$

The elegance of the numbers reveals itself. One of the $2.2$ terms in the numerator perfectly cancels with the denominator, leaving us with:

$$ P_{total} = 2.2	ext{ W} $$

And just like that, by carefully decoding the diagram and paying attention to the specific wording of the question, we have arrived at the flawless answer. Never let a complex diagram intimidate you; break it down into its fundamental parallel and series components!

Animated Solution for Physics - Current Electricity: In the given figure, the emf of the cell is $2.2 V$ and if internal resistance is $0.6Ω$ . Calculate the power dissipated in the whole circuit

Select Answer:

Visualized Solution

Circuit Analysis

Parallel Branches

Branch Resistances

Equivalent External Resistance

Total Power Formula

Substitution

Final Calculation

Key Takeaway

The Sigma Insight: Cells, EMF, and Internal Resistance

Decoding the Circuit Diagram

The Power of Parallelism

Calculating the Equivalent Resistance

The Whole Circuit Power

The Final Strike

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