a battery of three cells in series each of emf 3volt an internal resistances of 0.5 ohms is connected to 2 ohms resistors in series with a parallel combination of 2,3, ohms resistor [1] draw the digram and calculate [a] effective [b] the I in circuit [c] the lost voltage battery [d] the current in one of the 3 ohm resistors. [a] 3.5 ohm [b]1.2A [c] 1.8V [d] 0.6A
To answer the given questions, we need to follow these steps and utilize the given information:
Step 1: Drawing the circuit diagram:
In the given circuit, three cells with an EMF of 3 volts and an internal resistance of 0.5 ohms are connected in series. The series combination is then connected to a 2-ohm resistor in series with a parallel combination of three resistors: 2 ohms, 3 ohms, and another 2 ohms.
Here is the diagram of the circuit:
```
+---------+-----------+-------------+
| | |
| V |
| - |
+---------+
| |
| +--+--+
| [3V] | |
| | 0.5|
| +--+--+
| |
| |
| +-+
| | |
| [R1]
| [+]
+--------+---------+---------+
| | | |
+-+ V V V
| | - - -
[R2] [R3] [R4] [R5]
[+]
|
|
[+]
|
|
[R6]
[+]
```
Step 2: Calculating the effective resistance (a):
The total resistance in the circuit can be found by summing the resistances in series and parallel.
The series resistance of the three cells is: R_cell = 3 * 0.5 = 1.5 ohms.
The parallel combination of 2 ohms, 3 ohms, and 2 ohms can be simplified to a single equivalent resistance (Req) using the formula: Req = 1 / (1/R2 + 1/R3 + 1/R4).
Substituting the values: Req = 1 / (1/2 + 1/3 + 1/2) = 1 / (3/6 + 2/6 + 3/6) = 1 / (8/6) = 6/8 = 0.75 ohms.
The total resistance in the circuit is the sum of R_cell and Req: R_total = R_cell + Req = 1.5 + 0.75 = 2.25 ohms.
Therefore, the effective resistance in the circuit (a) is 2.25 ohms.
Step 3: Calculating the total current (b):
To calculate the total current (I) in the circuit, we can use Ohm's Law: V = I * R, where V is the EMF of the battery.
Given that the EMF (V) of the battery is 3 volts, and the effective resistance (R_total) is 2.25 ohms, we can substitute these values into the equation to find the total current:
I = V / R_total = 3 / 2.25 ≈ 1.33 A.
Therefore, the total current in the circuit (b) is approximately 1.33 A.
Step 4: Calculating the lost voltage across the battery (c):
The lost voltage across the battery can be calculated using the formula: V_lost = I * r, where r is the internal resistance of the battery cells.
Given that the internal resistance (r) is 0.5 ohms, and the total current (I) is 1.33 A, we can substitute these values into the equation to find the lost voltage:
V_lost = I * r = 1.33 * 0.5 ≈ 0.67 V.
Therefore, the lost voltage across the battery (c) is approximately 0.67 V.
Step 5: Calculating the current in one of the 3-ohm resistors (d):
Since the resistors in the parallel combination have the same voltage across them, we can calculate the current flowing through any resistor in that combination.
Using Ohm's Law: V = I * R, where V is the voltage across the resistor and R is the resistance.
The voltage across the resistors connected in parallel is equal to the EMF of the battery (V = 3 volts).
Using this voltage and the resistance of the 3-ohm resistor (R = 3 ohms), we can calculate the current flowing through it:
I = V / R = 3 / 3 = 1 A.
Therefore, the current in one of the 3-ohm resistors (d) is 1 A.
To summarize the answers:
[a] Effective Resistance: 2.25 ohms
[b] Total Current: 1.33 A
[c] Lost Voltage across the battery: 0.67 V
[d] Current in one of the 3-ohm resistors: 1 A.