Problem 144 - A Complete Resource Book in Physics for JEE Main

Question

A resistance of \(4 \Omega\) and a wire of length \(5 \mathrm{~m}\) and resistance \(5 \Omega\) are joined in series and connected to a cell of EMF \(10 \mathrm{~V}\) and internal resistance \(1 \Omega\). A parallel combination of two identical cells is balanced across \(300 \mathrm{~cm}\) of the wire. The EMF of each cell is

Step-by-Step Solution

Answer

The EMF of each cell in the parallel combination is \(1.5V\).

Step 1: Calculate the total resistance

First, let's calculate the total resistance of the circuit, which includes the 4Ω resistor, the 5Ω wire, and the internal resistance of the cell. Since they are connected in series, we can simply add up their resistances: Total resistance, R_total = R_resistor + R_wire + R_internal R_total = 4Ω + 5Ω + 1Ω = 10Ω

Step 2: Determine the current

Now that we know the total resistance of the circuit, we can use Ohm's law to determine the current flowing through it. Ohm's law states that Voltage (V) = Current (I) × Resistance (R). Rearranging to solve for the current, we get: I = V / R I = 10V / 10Ω = 1A The current flowing through the circuit is 1A.

Step 3: Calculate the potential difference across the 300cm wire section

Next, we need to find the potential difference across the 300cm section of the 5m wire (remembering to convert 300cm to meters). First, we determine the resistance of the 300cm section using the proportion: R_300cm = (Resistance of the wire) × (length of wire section) / (total length of the wire) R_300cm = 5Ω × (300cm / 100cm) / 5m = 3Ω Now we can use Ohm's law to calculate the potential difference across the 300cm section: V_300cm = I * R_300cm V_300cm = 1A * 3Ω = 3V The potential difference across the 300cm section of the wire is 3V.

Step 4: Find the EMF of each cell

Finally, we can find the EMF of each identical cell in the parallel combination. Since the cells are connected in parallel, the potential difference across them is the same. We have found the potential difference across the 300cm section of the wire, which is 3V. Since the cells are identical, their EMFs are equal, and the EMF of each cell can be represented by E: 3V = E + E 3V = 2E E = 1.5V Hence, the EMF of each cell is 1.5V.