2 cells, each of Emf 2v and internal resistance 0.5ohms are connected in series .they are made to supply current to a combination of three resistors one of the resistance 2ohms is connected in series to a parallel combination of two other resistor each of resistance 3ohms.draw the circuit diagram and Calcucate
1. Current in the circuit
2. Potential difference across the parallel combination of the resistor.
3.losts volt of the battery
Solve
Chiemerie
Answer please
To answer these questions, we need to analyze the given circuit. Let's start by drawing the circuit diagram:
Here is the circuit diagram:
```
_______ _______ _______
| | | | | |
----| 2V, |-----| 2V, |-----| 2V, |-----
| 0.5Ω | | 0.5Ω | | 0.5Ω |
----| |-----| |-----| |-----
|_______| |_______| |_______|
| |
| |
--| 2Ω |--
| |
|_______|
_______
| |
--| |--
_| 3Ω |_
| | | |
| |_______| |
| _______ |
| | | |
| | 3Ω | |
--|_______|--
```
Now, let's move on to calculating the values:
1. Current in the circuit:
To find the total current flowing in the circuit, we need to calculate the equivalent resistance of the combination of resistors connected in series.
Let's analyze the series circuit:
```
_______ _______ _______
| | | | | |
---- | 2V, |-----| 2V, |-----| 2V, |-----
| 0.5Ω | | 0.5Ω | | 0.5Ω |
---- | |-----| |-----| |-----
|_______| |_______| |_______|
```
The total resistance (R_total) in this series circuit is the sum of the resistances of the individual cells:
R_total = R1 + R2 = 0.5Ω + 0.5Ω = 1Ω
Now, to calculate the total current flowing in the circuit (I_total), we can use Ohm's Law:
I_total = Emf_total / R_total
Since the Emf (E) of each cell is given as 2V, the total Emf of the two cells in series (Emf_total) will be:
Emf_total = E1 + E2 = 2V + 2V = 4V
Now, substitute the values into the formula:
I_total = 4V / 1Ω = 4A
Therefore, the current in the circuit is 4A.
2. Potential difference across the parallel combination of the resistor:
To calculate the potential difference across the parallel combination of the resistors, we need to calculate the current flowing through the combination.
Let's analyze the parallel combination circuit:
```
| |
| |
--| 2Ω |--
| |
|_______|
_______
| |
--| |--
_| 3Ω |_
| | | |
| |_______| |
| _______ |
| | | |
| | 3Ω | |
--|_______|--
```
The total resistance (R_total) in this parallel circuit can be calculated using the formula:
1/R_total = 1/R1 + 1/R2 + 1/R3
Substituting the given resistances:
1/R_total = 1/2Ω + 1/3Ω + 1/3Ω
To simplify the calculation, we can find the least common denominator:
1/R_total = 3/6Ω + 2/6Ω + 2/6Ω
1/R_total = 7/6Ω
Now, to calculate the total current flowing through this parallel combination, we can use Ohm's Law. We will use the total potential difference (V_total) across the combination, which is equal to the Emf_total:
I_total_parallel = V_total / R_total
Since the total Emf (V_total) is given as 4V from the previous calculation, we can substitute the values into the formula:
I_total_parallel = 4V / (7/6Ω)
To divide by a fraction, we multiply by its reciprocal:
I_total_parallel = 4V * (6/7Ω)
Simplifying this calculation:
I_total_parallel = 24V / 7Ω
Therefore, the current flowing through the parallel combination of resistors is approximately 3.43A.
The potential difference across the parallel combination of the resistors is the same as the total potential difference (V_total), which is 4V.
3. Lost voltage of the battery:
The lost voltage (V_lost) is the potential difference across the internal resistance of the cells. We can calculate it using Ohm's Law:
V_lost = I_total * R_internal
Substituting the values:
V_lost = 4A * 0.5Ω
V_lost = 2V
Therefore, the lost voltage of the battery is 2V.