Electric circuit with three cells each of emf 2v and negligible internal resistances are connected in series with an ammeter two resistors of resistance 2 and 6 and a key.
1. Draw a diagram for the above information
2.calculate the current in the circuit when the key is closed
6 volt battery
resistance of 8 ohms
I assume the resistors are also in series
i = 6/8 = 3/4 = 0.75 amps
Answers
1. Here is a diagram representing the given electric circuit description:
```
_______ _______ _______
| | | | | |
| 2V | | 2V | | 2V |
| | | | | |
---|_______|--|_______|--|_______|---
| | | | | |
| 2Ω | | 6Ω | |Ammeter|
| | | | | |
---|_______|--|_______|--|_______|---
| Key |
|_____________________________|
```
2. To calculate the current in the circuit, we can use Ohm's Law, which states that the current (I) flowing through a circuit is equal to the total applied voltage (V) divided by the total resistance (R).
First, let's find the total resistance in the circuit:
- The resistors with resistances of 2Ω and 6Ω are connected in series, so their total resistance (R₁) is:
R₁ = 2Ω + 6Ω = 8Ω
- The total resistance (R) in series is simply the sum of all the resistances.
Next, let's find the total applied voltage in the circuit:
- Since the cells are connected in series, the total applied voltage (V) is the sum of the individual cell voltages:
V = 2V + 2V + 2V = 6V
- The total applied voltage (V) in series is simply the sum of all the cell voltages.
Now, we can use Ohm's Law to calculate the current (I):
- I = V / R
- I = 6V / 8Ω
- I = 0.75A
Therefore, the current in the circuit when the key is closed is 0.75A.