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Answer :
We start by finding the equivalent resistance of the first group, which consists of three resistors connected in parallel with resistances
[tex]$$
R_1 = 10\,\Omega,\quad R_2 = 20\,\Omega,\quad R_3 = 30\,\Omega.
$$[/tex]
For resistors in parallel, the equivalent resistance, [tex]$R_{\text{group1}}$[/tex], is given by
[tex]$$
\frac{1}{R_{\text{group1}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}.
$$[/tex]
Calculating this, we obtain an equivalent resistance of approximately
[tex]$$
R_{\text{group1}} \approx 5.4545\,\Omega.
$$[/tex]
Next, we consider the second group, where two resistors are connected in parallel with resistances
[tex]$$
R_4 = 3\,\Omega \quad \text{and} \quad R_5 = 6\,\Omega.
$$[/tex]
Their equivalent resistance, [tex]$R_{\text{group2}}$[/tex], is found using
[tex]$$
\frac{1}{R_{\text{group2}}} = \frac{1}{R_4} + \frac{1}{R_5}.
$$[/tex]
This results in
[tex]$$
R_{\text{group2}} \approx 2\,\Omega.
$$[/tex]
Now that we have the equivalent resistances of the two groups, we note that these groups are connected in series. The total resistance, [tex]$R_{\text{total}}$[/tex], is the sum of the two:
[tex]$$
R_{\text{total}} = R_{\text{group1}} + R_{\text{group2}} \approx 5.4545\,\Omega + 2\,\Omega \approx 7.4545\,\Omega.
$$[/tex]
Finally, using Ohm’s law with a battery voltage of [tex]$V = 45\,\text{V}$[/tex], the total current in the circuit is given by
[tex]$$
I = \frac{V}{R_{\text{total}}} = \frac{45}{7.4545} \approx 6.0366\,\text{A}.
$$[/tex]
Thus, the equivalent resistances and the current in the circuit are:
- First group equivalent resistance: [tex]$R_{\text{group1}} \approx 5.4545\,\Omega$[/tex]
- Second group equivalent resistance: [tex]$R_{\text{group2}} \approx 2\,\Omega$[/tex]
- Total resistance: [tex]$R_{\text{total}} \approx 7.4545\,\Omega$[/tex]
- Total current: [tex]$I \approx 6.0366\,\text{A}$[/tex]
[tex]$$
R_1 = 10\,\Omega,\quad R_2 = 20\,\Omega,\quad R_3 = 30\,\Omega.
$$[/tex]
For resistors in parallel, the equivalent resistance, [tex]$R_{\text{group1}}$[/tex], is given by
[tex]$$
\frac{1}{R_{\text{group1}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}.
$$[/tex]
Calculating this, we obtain an equivalent resistance of approximately
[tex]$$
R_{\text{group1}} \approx 5.4545\,\Omega.
$$[/tex]
Next, we consider the second group, where two resistors are connected in parallel with resistances
[tex]$$
R_4 = 3\,\Omega \quad \text{and} \quad R_5 = 6\,\Omega.
$$[/tex]
Their equivalent resistance, [tex]$R_{\text{group2}}$[/tex], is found using
[tex]$$
\frac{1}{R_{\text{group2}}} = \frac{1}{R_4} + \frac{1}{R_5}.
$$[/tex]
This results in
[tex]$$
R_{\text{group2}} \approx 2\,\Omega.
$$[/tex]
Now that we have the equivalent resistances of the two groups, we note that these groups are connected in series. The total resistance, [tex]$R_{\text{total}}$[/tex], is the sum of the two:
[tex]$$
R_{\text{total}} = R_{\text{group1}} + R_{\text{group2}} \approx 5.4545\,\Omega + 2\,\Omega \approx 7.4545\,\Omega.
$$[/tex]
Finally, using Ohm’s law with a battery voltage of [tex]$V = 45\,\text{V}$[/tex], the total current in the circuit is given by
[tex]$$
I = \frac{V}{R_{\text{total}}} = \frac{45}{7.4545} \approx 6.0366\,\text{A}.
$$[/tex]
Thus, the equivalent resistances and the current in the circuit are:
- First group equivalent resistance: [tex]$R_{\text{group1}} \approx 5.4545\,\Omega$[/tex]
- Second group equivalent resistance: [tex]$R_{\text{group2}} \approx 2\,\Omega$[/tex]
- Total resistance: [tex]$R_{\text{total}} \approx 7.4545\,\Omega$[/tex]
- Total current: [tex]$I \approx 6.0366\,\text{A}$[/tex]
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