a battery of three cell is series each of emf two volt and internal resistant 0.5 ohms is connected to a two ohm resistor in series with a parallel combination of 2.3 ohms resistor draw the circuit diagram and calculate
effective external resistance
effective resistance in the circuit
current in the circuit
lost volt in the circuit
current in one of the three ohms resistor
See Related Questions: Tue, 11-23-2010, 5:05 AM.
To draw the circuit diagram, you can follow these steps:
1. Draw a straight line to represent the circuit layout.
2. Place three cells in series, connecting the positive terminal of one cell to the negative terminal of the next cell.
3. Draw a resistor symbol for the 2 Ohm resistor connected in series with the cells.
4. Place a branch in the circuit for the parallel combination of the 2.3 Ohm resistor.
5. Connect the remaining end of the 2.3 Ohm resistor to the end of the 2 Ohm resistor that is opposite to the cells.
The resulting circuit diagram will look like this:
```plaintext
[ 2V | 0.5Ω ] -- 2Ω -- [ 2.3Ω | 2.3Ω | 2.3Ω ]
```
Now, let's calculate the requested values:
1. Effective External Resistance:
The effective external resistance (R_eff) can be found by adding the resistances in series. In this case, R_eff = 2 Ω + 2.3 Ω = 4.3 Ω.
2. Effective Resistance in the Circuit:
The effective resistance in the circuit (R_total) is the sum of the external resistance (R_eff) and the internal resistance (r) of the battery. In this case, R_total = R_eff + r = 4.3 Ω + 0.5 Ω = 4.8 Ω.
3. Current in the Circuit:
To calculate the current (I) in the circuit, we can apply Ohm's Law. Ohm's Law states that I = V / R, where V is the total voltage and R is the total resistance.
In this case, the total voltage (V) is the sum of the individual cell voltages, which is 2V x 3 = 6V. Therefore, I = 6V / 4.8 Ω = 1.25 A.
4. Lost Voltage in the Circuit:
The lost voltage in the circuit is the voltage drop across the internal resistance of the battery. Using Ohm's Law (V = I * r), where r is the internal resistance of each cell (0.5 Ω), the lost voltage can be calculated as V_lost = I * r = 1.25 A * 0.5 Ω = 0.625 V.
5. Current in One of the Three-Ohm Resistors:
Since the three-ohm resistors are connected in parallel, they have the same voltage across them. Therefore, the current flowing through each of the three-ohm resistors would be I = V / R, where V is the voltage across the resistors and R is the resistance of each resistor (3 Ω). From the circuit diagram, the voltage across the three-ohm resistors is the same as the total voltage, which is 6V. Thus, the current in one of the three-ohm resistors would be I = 6V / 3 Ω = 2 A.