1. An electrical circuit consists of a battery with 2 cells,each with an EMF of 12 volts and an internal resistance of of 0.4 ohms,connected in parallel. An ammeter with an internal resistance of 0.15 ohms is connected in series with resistor Ra=2ohms and Rb=4ohms
Calculate the following
A. The total internal resistance of the battery.
B. The ammeter reading
A. Ri = 0.4 + 0.4 = 0.8 Ohms
B. I = E/(Ri+Rm+Ra+Rb) = 24/6.95 = 3.45A
A. To calculate the total internal resistance of the battery, we need to consider the internal resistances of both cells. Since the cells are connected in parallel, the total internal resistance is given by the reciprocal of the sum of the reciprocals of the individual internal resistances.
The formula for the total internal resistance (R_total) of two resistors in parallel is:
1/R_total = 1/R1 + 1/R2
In this case, R1 = R2 = 0.4 ohms.
1/R_total = 1/0.4 + 1/0.4
Simplifying,
1/R_total = 2.5 + 2.5
1/R_total = 5
Therefore,
R_total = 1/5 = 0.2 ohms
So, the total internal resistance of the battery is 0.2 ohms.
B. To calculate the ammeter reading, we need to consider the total resistance in the circuit. The ammeter is connected in series with resistors Ra and Rb, as well as the internal resistance of the ammeter itself.
The total resistance (R_total) in the circuit is given by the sum of the individual resistances:
R_total = Ra + Rb + internal resistance of the ammeter
In this case, Ra = 2 ohms, Rb = 4 ohms, and the internal resistance of the ammeter = 0.15 ohms.
R_total = 2 + 4 + 0.15
R_total = 6.15 ohms
Now, to calculate the ammeter reading, we can use Ohm's Law, which states that the current (I) flowing through a circuit is given by the following formula:
I = V / R
Where V is the total voltage in the circuit and R is the total resistance.
In this case, the total voltage is given by the EMF of the battery, which is the same for both cells and equals 12 volts.
So,
I = 12 / 6.15
Calculating,
I ≈ 1.951 amps
Therefore, the ammeter reading is approximately 1.951 amps.