Answer :
To solve the problem, we first determine the effective resistance of each group of resistors and then find the total current from the battery.
1. Resistors in Group 1 (resistors of [tex]$18\,\Omega$[/tex], [tex]$20\,\Omega$[/tex], and [tex]$30\,\Omega$[/tex]) are connected in parallel. The formula for the equivalent resistance [tex]$R_{eq1}$[/tex] of resistors in parallel is given by
[tex]$$
\frac{1}{R_{eq1}} = \frac{1}{18} + \frac{1}{20} + \frac{1}{30}.
$$[/tex]
Calculating the sum of the reciprocals:
[tex]$$
\frac{1}{18} + \frac{1}{20} + \frac{1}{30} \approx 0.05556 + 0.05 + 0.03333 \approx 0.13889,
$$[/tex]
so
[tex]$$
R_{eq1} = \frac{1}{0.13889} \approx 7.2\,\Omega.
$$[/tex]
2. Resistors in Group 2 (resistors of [tex]$3\,\Omega$[/tex] and [tex]$6\,\Omega$[/tex]) are also connected in parallel. Their equivalent resistance [tex]$R_{eq2}$[/tex] is found using
[tex]$$
\frac{1}{R_{eq2}} = \frac{1}{3} + \frac{1}{6}.
$$[/tex]
Computing the reciprocals:
[tex]$$
\frac{1}{3} + \frac{1}{6} \approx 0.33333 + 0.16667 \approx 0.5,
$$[/tex]
thus
[tex]$$
R_{eq2} = \frac{1}{0.5} = 2\,\Omega.
$$[/tex]
3. The two groups are then connected in series. For resistors in series, the total resistance [tex]$R_{total}$[/tex] is the sum of the individual resistances:
[tex]$$
R_{total} = R_{eq1} + R_{eq2} \approx 7.2\,\Omega + 2\,\Omega = 9.2\,\Omega.
$$[/tex]
4. Finally, the battery voltage is [tex]$45\,V$[/tex]. Using Ohm's law, the total current [tex]$I$[/tex] drawn from the battery is given by
[tex]$$
I = \frac{V}{R_{total}} = \frac{45}{9.2} \approx 4.8913\,\text{A}.
$$[/tex]
Thus, the effective resistances of the two groups are approximately [tex]$7.2\,\Omega$[/tex] and [tex]$2\,\Omega$[/tex], with a total resistance of [tex]$9.2\,\Omega$[/tex], and the current from the battery is approximately [tex]$4.8913\,\text{A}$[/tex].
1. Resistors in Group 1 (resistors of [tex]$18\,\Omega$[/tex], [tex]$20\,\Omega$[/tex], and [tex]$30\,\Omega$[/tex]) are connected in parallel. The formula for the equivalent resistance [tex]$R_{eq1}$[/tex] of resistors in parallel is given by
[tex]$$
\frac{1}{R_{eq1}} = \frac{1}{18} + \frac{1}{20} + \frac{1}{30}.
$$[/tex]
Calculating the sum of the reciprocals:
[tex]$$
\frac{1}{18} + \frac{1}{20} + \frac{1}{30} \approx 0.05556 + 0.05 + 0.03333 \approx 0.13889,
$$[/tex]
so
[tex]$$
R_{eq1} = \frac{1}{0.13889} \approx 7.2\,\Omega.
$$[/tex]
2. Resistors in Group 2 (resistors of [tex]$3\,\Omega$[/tex] and [tex]$6\,\Omega$[/tex]) are also connected in parallel. Their equivalent resistance [tex]$R_{eq2}$[/tex] is found using
[tex]$$
\frac{1}{R_{eq2}} = \frac{1}{3} + \frac{1}{6}.
$$[/tex]
Computing the reciprocals:
[tex]$$
\frac{1}{3} + \frac{1}{6} \approx 0.33333 + 0.16667 \approx 0.5,
$$[/tex]
thus
[tex]$$
R_{eq2} = \frac{1}{0.5} = 2\,\Omega.
$$[/tex]
3. The two groups are then connected in series. For resistors in series, the total resistance [tex]$R_{total}$[/tex] is the sum of the individual resistances:
[tex]$$
R_{total} = R_{eq1} + R_{eq2} \approx 7.2\,\Omega + 2\,\Omega = 9.2\,\Omega.
$$[/tex]
4. Finally, the battery voltage is [tex]$45\,V$[/tex]. Using Ohm's law, the total current [tex]$I$[/tex] drawn from the battery is given by
[tex]$$
I = \frac{V}{R_{total}} = \frac{45}{9.2} \approx 4.8913\,\text{A}.
$$[/tex]
Thus, the effective resistances of the two groups are approximately [tex]$7.2\,\Omega$[/tex] and [tex]$2\,\Omega$[/tex], with a total resistance of [tex]$9.2\,\Omega$[/tex], and the current from the battery is approximately [tex]$4.8913\,\text{A}$[/tex].
