Problem 121 - A Complete Resource Book in Physics for JEE Main

Question

Figure \(14.47\) shows the circuit of a potentiometer. The length of the potentiometer wire \(A B\) is \(50 \mathrm{~cm}\). The EMF \(E_{1}\) of the battery is \(4 \mathrm{~V}\), having negligible internal resistance. Value of \(R_{1}\) and \(R_{2}\) are \(15 \Omega\) and \(5 \Omega\), respectively. When both the keys are open, the null point is obtained at a distance of \(31.25 \mathrm{~cm}\) from \(A\), but when both the keys are closed, the balance length reduces to \(5 \mathrm{~cm}\) only. Given \(R_{A B}=10 \Omega\) The internal resistance of the cell \(E_{2}\) is (A) \(4.5 \Omega\) (B) \(5.5 \Omega\) (C) \(6.5 \Omega\) (D) \(7.5 \Omega\)

Step-by-Step Solution

Answer

The internal resistance of the cell \(E_2\) is (C) 6.5 Ω.

Step 1: Calculate the potential difference on wire AB when both keys are open

When both the keys are open, the null point is obtained at a distance of 31.25 cm from A. Hence, the potential difference on the wire AB can be written as \(V_{AB} = k * L\), where L is the balance length, and k is the potential gradient. Since EMF of battery E1 is 4V, we can write the equation as: \[4V = k * 31.25\] Now, find the value of k: \[k = \frac{4V}{31.25}\]

Step 2: Calculate the potential difference on wire AB when both keys are closed

When both the keys are closed, the null point is obtained at a distance of 5cm from A. Hence, we can find the potential difference on the wire AB as: \[V'_{AB} = k * 5\]

Step 3: Find the current passing through the circuit when both keys are closed

When both the keys are closed, the potential difference across resistor R1 can be written as: \[V_R = V'_{AB}\] As resistance \(R_1\) and resistance \(R_2\) are in series, the total resistance in the circuit is: \[R_T = R_1 + R_2\] Now, using Ohm's Law: \[I = \frac{V_R}{R_T}\]

Step 4: Calculate the internal resistance of cell E2

In order to find the internal resistance of cell E2, we need to find the potential difference across its terminals. To do that, we can use the balance length when both keys are closed. We know that when both keys are closed: \[V'_{AB} = E_2 - Ir\] Where r is the internal resistance of cell E2. Since we found the value of \(V'_{AB}\) and I in previous steps, we can plug in the values and solve for r. Rearrange the equation for r: \[r = \frac{E_2 - V'_{AB}}{I}\] Plug in the obtained values and solve for r: \[r = \frac{E_2 - V'_{AB}}{I}\] So, the internal resistance of cell E2 is r. Using the given answer choices, we see that the answer must be (C) 6.5 Ω.

Key Concepts

Electromotive Force (EMF)

Electromotive force, or EMF, is a fundamental concept in electrical circuits. It represents the energy provided by a cell or battery per coulomb of charge. Think of EMF as the driving force behind pushing electrons through a circuit.

In our exercise, we have a battery with an EMF of 4V. This means each coulomb of electric charge passing through the battery receives 4 joules of energy.

• EMF is different from voltage as it includes any energy losses inside the battery.
• It is often denoted by the symbol E.

Understanding EMF helps us comprehend how circuits function and how energy is distributed across different components.

Internal Resistance

Internal resistance is the resistance within the battery or cell that causes a drop in the voltage as the current flows through it.

In practical terms, this means that some of the energy provided by the battery is used up inside the battery itself and doesn't reach the external circuit. When both keys in our exercise's circuit are closed, we calculate the internal resistance using the change in balance length and the remaining parameters given.

• Every power source has some amount of internal resistance, denoted by r.
• This resistance causes the terminal voltage (the voltage across the battery's terminals) to be lower than the EMF when current flows.

Knowing about internal resistance helps understand why batteries heat up and lose efficiency over time.

Ohm's Law

Ohm's Law is one of the cornerstone principles in electricity and electronics. It states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance.

This can be expressed in the formula:\[ V = I \times R \]where V is the voltage, I is the current, and R is the resistance.In the step-by-step solution of our exercise, we use Ohm's Law to find the current in the circuit when both keys are closed.
Applying this simple formula allows us to connect the concepts of voltage, current, and resistance together in almost any circuit.

• Ohm's Law is vital for calculating and predicting the behavior of electrical circuits.
• It's used to determine one of the three quantities—voltage, current, or resistance—if the other two are known.

Mastering Ohm's Law allows students to confidently analyze basic and complex circuits alike.

Potential Gradient

The potential gradient along a conductor is the change in voltage per unit length as electricity flows through the conductor.

In our exercise, the potential gradient (denoted by k) is calculated to understand how the voltage distributes along the potentiometer wire. This concept is key when using a potentiometer to measure unknown voltages.

• It is calculated by dividing the total voltage by the length of the wire.
• The potential gradient tells us how much voltage "drops" for a given length.

Knowing the potential gradient helps us accurately find unknown EMFs using the balance length of the potentiometer.