Question

when a cell is connected to a resistance \(R_{1}\) the rate at which heat is generated in it is the same as when the cell is connected to a resistance \(\mathrm{R}_{2}\left(<\mathrm{R}_{1}\right)\) the internal resistance of the cell is.... (A) \(\left(R_{1}-R_{2}\right)\) (B) \((1 / 2)\left(\mathrm{R}_{1}-\mathrm{R}_{2}\right)\) (C) \(\left\\{\left(\mathrm{R}_{1} \mathrm{R}_{2}\right) /\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)\right.\) (D) \(\sqrt{R}_{1} R_{2}\)

Short answer

The internal resistance of the cell is: \(r = \cfrac{R_1 R_2}{R_1 + R_2}\)

Step-by-step solution

Establish relationships for power in both cases

In case 1, the cell is connected to resistance R₁, and in case 2, it is connected to resistance R₂. Let's denote the internal resistance of the cell as r, the total resistance of the circuit in case 1 as R₁' and in case 2 as R₂'. Then: \(R_1' = R_1 + r\) \(R_2' = R_2 + r\) The power in each situation, using the formula P = I²R is: \(P_1 = I_1^2 R_1\) \(P_2 = I_2^2 R_2\) Since the problem states that the rate of heat generation in both cases is the same, we can set the power equal to each other. \(P_1 = P_2\) Now we have two equations representing the power in each case. We need to find a relationship between the currents and resistances to determine the internal resistance r.

Find current in terms of resistances in both cases

We can use Ohm's law, V = IR, to find the current in each case. For case 1, we have: \(I_1 = \cfrac{E}{R_1'}\), where E is the EMF of the cell. Similarly, for case 2, we have: \(I_2 = \cfrac{E}{R_2'}\) Now we can write the power equations in terms of resistances: \(P_1 = \left(\cfrac{E}{R_1'}\right)^2 R_1\) \(P_2 = \left(\cfrac{E}{R_2'}\right)^2 R_2\) Since P₁ = P₂: \(\left(\cfrac{E}{R_1'}\right)^2 R_1 = \left(\cfrac{E}{R_2'}\right)^2 R_2\)

Solve for internal resistance r

Since E is nonzero, we can divide both sides by E²: \(\cfrac{R_1}{R_1'^2} = \cfrac{R_2}{R_2'^2}\) Now substitute the expressions for total resistances found in Step 1: \(\cfrac{R_1}{(R_1 + r)^2} = \cfrac{R_2}{(R_2 + r)^2}\) To solve for r, cross-multiply and simplify: \(R_1 (R_2 + r)^2 = R_2 (R_1 + r)^2\) Expanding the squares and simplifying further, we have: \(R_1 R_2^2 + 2 R_1 R_2 r + R_1 r^2 = R_2 R_1^2 + 2 R_2 R_1 r + R_2 r^2\) Canceling equal terms and solving for r, we get: \(r = \cfrac{R_1 R_2}{R_1 + R_2}\) The correct answer is: (C) \(\left\\{\left(\mathrm{R}_{1} \mathrm{R}_{2}\right) /\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)\right.)$

Key concepts

Ohm's Law

Ohm's Law is a fundamental principle in electronics and electrical engineering. It tells us about the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit. This law can be simply put as:

• Voltage (V) = Current (I) × Resistance (R)

This equation implies that the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance of the conductor.
In essence, if you increase the voltage in the circuit while keeping the resistance constant, the current will increase proportionally. Alternatively, increasing resistance while maintaining the same voltage will reduce the current.

For calculations: use Ohm's law to relate the quantities when dealing with internal resistance calculations.

Electromotive Force (EMF)

Electromotive force, or EMF, refers to the energy provided by a power source to move electrons through a circuit. Although EMF is measured in volts, it represents the potential difference across the terminals of a battery or generator when no current is flowing.
A key distinction is that EMF is not a "force" in the traditional sense; rather, it's a measure of energy provided per coulomb of charge. It's what "pushes" the electrons and allows the circuit's current to flow.

• In the context of the exercise, EMF is crucial because it’s the driving voltage behind the power dissipation in the resistors.
• When the cell is connected to a resistance, EMF needs to account for internal resistance to correctly describe the distribution of voltage in the circuit.

Power Dissipation

Power dissipation refers to the conversion of electrical energy into heat energy in a circuit. This is a natural consequence when current flows through a resistive element. Power dissipation can be calculated using the formula:

• P = I²R, where P is power, I is current, and R is resistance.

This equation indicates that power dissipation is proportional to both the square of the current and the resistance.
In our exercise, power dissipation was used to equate two scenarios where a cell was connected to two different resistors, highlighting the balance between internal resistance and load.

• This understanding is crucial for determining the balance in the circuit, which in turn helps calculate unknown quantities such as internal resistance.

Resistors in Series

When resistors are connected end-to-end in a circuit, they are said to be in series. This setup affects the total resistance, which is the sum of each individual resistance:

• R_{total} = R_1 + R_2 + ... + R_n

This means that the current flowing through each resistor is the same, but the total voltage across the circuit is divided among them.
In the context of this exercise, when determining the internal resistance of the cell, the concept of resistors in series helps deduce how the internal resistance 'adds' to the load resistance. This relationship becomes crucial in satisfying the condition of equal power dissipation between two configurations of resistors linked in series with the cell.
Understanding these relationships then allows students to solve for internal resistance based on given conditions.