To draw the circuit diagram, follow these steps:
1. Place three cells in a series. Connect the positive terminal of the first cell to the negative terminal of the second cell, and the positive terminal of the second cell to the negative terminal of the third cell.
2. Assign each cell an electromotive force (e.m.f) of 2V and an internal resistance of 0.5 ohms.
3. Connect a 2 ohm resistor in series with the battery arrangement from step 1.
4. In parallel to the 2 ohm resistor, connect two 3 ohm resistors.
The resulting circuit diagram should look like this:
[2V, 0.5Ω]----[2V, 0.5Ω]----[2V, 0.5Ω]
| |
| |
| |
| 2Ω Resistor |
| |
----[3Ω]----[3Ω]-----
Now, let's calculate the given parameters:
a) Effective External Resistance:
To find the effective external resistance, we need to calculate the total resistance of the parallel combination of the two 3 ohm resistors.
The formula for the total resistance of two resistors in parallel is given by:
1/Re = 1/R1 + 1/R2
Substituting the given values:
1/Re = 1/3 + 1/3 = 2/3
Re = 3/2 = 1.5Ω
Hence, the effective external resistance is 1.5Ω.
b) Current in the Circuit:
To find the current in the circuit, we can use Ohm's Law:
I = V / R
The total voltage (V) across the external resistance can be calculated as follows:
V = Total e.m.f. - Total internal resistance
V = 2V + 2V + 2V - (0.5Ω + 0.5Ω + 0.5Ω) = 6V - 1.5Ω = 4.5V
Now, substituting the values in the formula:
I = V / Re = 4.5V / 1.5Ω = 3A
Therefore, the current in the circuit is 3A.
c) Volts in Battery:
The voltage across each cell will be its e.m.f minus its internal resistance. So, the voltage across each cell would be 2V - 0.5Ω = 1.5V.
Therefore, the volts in the battery are 1.5V.
d) Current in one of the 3 ohm resistors:
As the 3 ohm resistors are connected in parallel, they have the same voltage across them.
Using Ohm's Law:
I = V / R = 1.5V / 3Ω = 0.5A
Therefore, the current in one of the 3 ohm resistors is 0.5A.