Three cells each of e.m.f. 1.5v and an internal resistance of \(1.0\ \Omega\) are connected in parallel across a load resistance of \(2.67\ \Omega\). Calcul...

Answer Details

The circuit given consists of three cells connected in parallel across a load resistance of 2.67 Ω. Each cell has an emf of 1.5V and an internal resistance of 1.0 Ω. To calculate the current in the load, we can use the formula: I = V / R where I is the current in the load, V is the total voltage across the load, and R is the resistance of the load. First, we need to calculate the total voltage across the load. Since the cells are connected in parallel, the voltage across each cell is the same as the total voltage, which is: V = 1.5V Next, we can use the formula for calculating the equivalent resistance of cells connected in parallel, which is: 1/R = 1/R1 + 1/R2 + 1/R3 where R1, R2, and R3 are the resistances of each cell. Since all cells have the same resistance, we can simplify the equation to: 1/R = 3/1 R = 1/3 Ω The total resistance of the circuit, including the internal resistance of the cells and the load resistance, is: Rt = R + r where r is the internal resistance of each cell, and Rt is the total resistance. So, Rt = 1/3 + 1 = 4/3 Ω Now, we can use Ohm's law to calculate the current in the load: I = V / Rt I = 1.5V / (4/3) Ω I = 0.5A Therefore, the current in the load is 0.50A.