Answered

A battery of three cells in series, each with an emf of [tex]$2 \, \text{V}$[/tex] and an internal resistance of [tex]$0.5 \, \Omega$[/tex], is connected to a [tex]$2 \, \Omega$[/tex] resistor in series with a parallel combination of two [tex]$3 \, \Omega$[/tex] resistors. Draw the circuit diagram and calculate:

a) The effective external resistance.
b) The current in the circuit.
c) The lost volts in the battery.
d) The current in one of the [tex]$3 \, \Omega$[/tex] resistors.



Answer :

GinnyAnswer
Let's solve the problem step-by-step.

### 1. Circuit Diagram

First, we draw the circuit diagram:

```
_____________ _______
| Cell (2V) | | |
| r = 0.5Ω | ---| |--- 2Ω resistor ---
|___________ | | |
(Repeats 2 more times to form a series with the above (two 3Ω resistors in parallel)
_______________

```

Here, three identical cells each of 2V and 0.5Ω are in series, producing a total emf and total internal resistance. This combined battery is connected in series to a 2Ω resistor and a parallel combination of two 3Ω resistors.

### Step-by-Step Solution:

### (a) Effective External Resistance

We need to first calculate the effective resistance of the 3Ω resistors in parallel:

For resistors in parallel:
[tex]\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} \][/tex]
Given [tex]\(R_1 = R_2 = 3Ω\)[/tex]:
[tex]\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \][/tex]
Thus,
[tex]\[ R_{\text{parallel}} = \frac{3}{2} = 1.5Ω \][/tex]

Now, the effective external resistance ([tex]\(R_{\text{ext}}\)[/tex]) is the series combination of the 2Ω resistor and the parallel combination of resistors:
[tex]\[ R_{\text{ext}} = 2Ω + 1.5Ω = 3.5Ω \][/tex]

### (b) Current in the Circuit

The total emf ([tex]\(E\)[/tex]) of the battery is the sum of the individual emfs:
[tex]\[ E = 2V + 2V + 2V = 6V \][/tex]

The total internal resistance ([tex]\(r\)[/tex]) is the sum of the individual internal resistances:
[tex]\[ r = 0.5Ω + 0.5Ω + 0.5Ω = 1.5Ω \][/tex]

The total resistance ([tex]\(R_{\text{total}}\)[/tex]) in the circuit is the sum of total internal resistance and the effective external resistance:
[tex]\[ R_{\text{total}} = r + R_{\text{ext}} = 1.5Ω + 3.5Ω = 5.0Ω \][/tex]

Using Ohm's law, the current ([tex]\(I\)[/tex]) in the circuit is:
[tex]\[ I = \frac{E}{R_{\text{total}}} = \frac{6V}{5.0Ω} = 1.2A \][/tex]

### (c) Lost Volts in the Battery

The voltage lost in the internal resistance ([tex]\(V_{\text{lost}}\)[/tex]) is given by:
[tex]\[ V_{\text{lost}} = I \times r = 1.2A \times 1.5Ω = 1.8V \][/tex]

### (d) Current in One of the 3Ω Resistors

The current through each resistor in parallel is equal to the total current divided by the number of parallel paths. Since the resistors are identical, the current is split equally:
[tex]\[ I_{\text{resistor}} = \frac{I}{2} = \frac{1.2A}{2} = 0.6A \][/tex]

### Summary:
- (a) The effective external resistance is [tex]\(3.5Ω\)[/tex].
- (b) The current in the circuit is [tex]\(1.2A\)[/tex].
- (c) The lost volts in the battery is [tex]\(1.8V\)[/tex].
- (d) The current in one of the 3Ω resistors is [tex]\(0.6A\)[/tex].

Other Questions