A battery of three cells in series, each of e.m.f 2V and internal resistance 0.5 Ω is connected to a 2 Ω resistor in series with a parallel combination of two 3 Ω resistor. Draw the circuit diagram and calculate:
1) the effective external resistance
2) the current in the circuit
3) the lost volts in the battery
4) the current in one of the 3 Ω resistors.
the two 3Ω resistors in parallel have an equivalent resistnce of 1.5Ω
So,
#1. 2 + 1.5 = 3.5Ω external R
See what you can do with that. It should not be too hard now.
To draw the circuit diagram, we'll start by representing the three cells in series as three vertical lines connected end to end with their positive (+) terminals at the top and negative (-) terminals at the bottom. Each cell has an electromotive force (emf) of 2V and an internal resistance of 0.5 Ω.
We'll label the positive terminal of the first cell as A, the negative terminal of the first cell as B, the positive terminal of the second cell as C, the negative terminal of the second cell as D, the positive terminal of the third cell as E, and the negative terminal of the third cell as F.
Next, we'll draw a resistor of 2 Ω in series with a parallel combination of two resistors of 3 Ω each. We'll label the two 3 Ω resistors as G and H. The other end of the 2 Ω resistor is connected to point B, and the other ends of the G and H resistors are connected to point F.
Now, let's move on to the calculations:
1) To find the effective external resistance, we need to find the total resistance of the circuit excluding the internal resistance of the batteries. In this case, it is the sum of the 2 Ω resistor and the parallel combination of the two 3 Ω resistors.
We can calculate the total resistance of the parallel combination of two resistors using the formula:
1/Re = 1/R1 + 1/R2
where Re is the effective resistance and R1, R2 are the resistances of the two resistors. Plugging in the values, we get:
1/Re = 1/3 + 1/3 = 2/3
Re = 3/2 Ω
The effective external resistance is found by adding the resistance of the 2 Ω resistor to the total resistance of the parallel combination:
Effective external resistance = 2 Ω + (3/2) Ω = 7/2 Ω
2) To find the current in the circuit, we can use Ohm's Law,
I = V/R,
where I is the current, V is the total voltage, and R is the effective external resistance. We already know the value of R from the previous calculation.
To find the total voltage, we can sum up the emf of the three cells. Each cell has an emf of 2V, so the total voltage is:
Total voltage = 2V + 2V + 2V = 6V
Plugging in the values, we get:
I = 6V / (7/2) Ω = 12/7 A
3) To find the lost volts in the battery, we can use the formula:
Lost volts = Internal resistance * Current
The internal resistance of each cell is given as 0.5 Ω, and we already know the current from the previous calculation.
Lost volts = 0.5 Ω * (12/7 A) = 6/7 V
4) To find the current in one of the 3 Ω resistors, we can use Ohm's Law again. The current flowing through the 3 Ω resistors is the same as the current flowing through the parallel combination of resistors, which we calculated to be 12/7 A.