In the circuit shown in figure the heat produced in the 5 ohm resistor due to the current flowing th

Q: Why do we use $P = i^2R$ instead of $P = V^2/R$?

Both formulas are perfectly valid! However, since we easily deduced the ratio of the currents in the two branches, using $P = i^2R$ is much faster and more direct. If you wanted to use $V^2/R$, you would first need to calculate the voltage across the parallel combination.

Q: Why is the current in the upper branch exactly half of the lower branch?

In a parallel circuit, current takes the path of least resistance. The upper branch has a total resistance of $10\,\Omega$, which is twice as much as the lower branch's $5\,\Omega$. Because the resistance is double, the current flowing through it will be exactly half.

Q: Does the $6\,\Omega$ resistor affect the heat produced in the $4\,\Omega$ resistor?

Absolutely! The $6\,\Omega$ resistor adds to the total resistance of the upper branch. This total resistance determines how much current flows through the entire upper branch. If the $6\,\Omega$ resistor weren't there, the current would be different, and so would the heat in the $4\,\Omega$ resistor.

Q: What if the $4\,\Omega$ and $6\,\Omega$ resistors were in parallel instead of series?

If they were in parallel, you would first calculate their equivalent parallel resistance. This new resistance would change the total resistance of the upper branch, which would completely alter how the current splits at the junction.

Q: Are calories per second a standard unit for electrical power?

Yes! Power is simply energy transferred per unit time. While we often use Watts (Joules per second) in electricity, calories per second is also a valid unit of power. The conversion is $1 \text{ cal/s} \approx 4.184 \text{ W}$.

Question

In the circuit shown in figure the heat produced in the $5 \Omega$ resistor due to the current flowing through it is $10 	ext{ cal/s}$.

The heat generated in the $4 \Omega$ resistor is

Options:

(A) 1 cal/s

(B) 2 cal/s

(C) 3 cal/s

(D) 4 cal/s

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are looking at a river that splits into two smaller streams before joining back together. This is exactly what is happening in our electrical circuit! The current enters from the left and faces a junction. It must decide how much of it will flow through the upper branch and how much will flow through the lower branch.

The lower branch is straightforward—it has a single $5\,\Omega$ resistor. The upper branch, however, has two resistors connected in series: a $4\,\Omega$ resistor and a $6\,\Omega$ resistor.

### The Current Split
To understand how the current divides, we first need to find the total resistance of the upper branch. Since the resistors are in series, we simply add them up:
$$R_{	ext{upper}} = 4\,\Omega + 6\,\Omega = 10\,\Omega$$

Now, here is the beautiful part about parallel circuits: the potential difference ($V$) across both branches is exactly the same. According to Ohm's Law ($V = iR$), if the voltage is constant, the current is inversely proportional to the resistance ($i \propto \frac{1}{R}$).

Notice that the upper branch has a resistance of $10\,\Omega$, which is exactly double the resistance of the lower branch ($5\,\Omega$). Because it has twice the resistance, it will only allow half as much current to flow through! If we let the current in the lower branch be $i$, the current in the upper branch will be $\frac{i}{2}$.

### The Power Equation
The problem gives us the heat produced in the $5\,\Omega$ resistor, which is $10 	ext{ cal/s}$. The formula for electrical power (or heat generated per second) is:
$$P = i^2R$$

For the lower branch, we can write:
$$P_5 = i^2 	imes 5 = 10 	ext{ cal/s}$$

Now, let's apply the same formula to the $4\,\Omega$ resistor in the upper branch. Remember, the current flowing through it is $\frac{i}{2}$:
$$P_4 = \left(\frac{i}{2}ight)^2 	imes 4$$

### Final Calculation
Let's simplify the expression for $P_4$:
$$P_4 = \frac{i^2}{4} 	imes 4 = i^2$$

We already know from the lower branch that $5i^2 = 10$, which means $i^2 = 2$.

Substituting this value into our equation for $P_4$, we get:
$$P_4 = 2 	ext{ cal/s}$$

And there we have it! The heat generated in the $4\,\Omega$ resistor is exactly **$2 	ext{ cal/s}$**.

Animated Solution for Physics - Current Electricity: In the circuit shown in figure the heat produced in the $5Ω$ resistor due to the current flowing through it is $10 cal/s$ . The heat generated in the $4Ω$ resistor is

Select Answer:

Visualized Solution

\text{Circuit Analysis}

R_{\text{upper}} = R_1 + R_2

i \propto \frac{1}{R}

i_{\text{upper}} = \frac{i_{\text{lower}}}{2}

P = i^2R

P_4 = \left(\frac{i}{2}\right)^2 \times 4

P_4 = \frac{P_5}{5}

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

Analyzing the Setup Imagine you are looking at a river that splits into two smaller streams before joining back together

The Current Split To understand how the current divides, we first need to find the total resistance of the upper branch

The Power Equation The problem gives us the heat produced in the $5 Ω$ resistor, which is $10 cal/s$

Final Calculation

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