Answer:
The current through the bell is approximately 2.8βA and the current through the lamp is approximately 5.25βA.
Explanation:
When resistors are connected in parallel, the reciprocal of the equivalent resistance ( [tex]R_{eq}[/tex] ) is equal to the sum of the reciprocals of the individual resistances. The formula for resistances in parallel is given by:
[tex]\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + ... + \frac{1}{R_{n}}[/tex]
In this case, with a 15-ohm bell ( [tex]R_{1}[/tex] β) and an 8.0-ohm lamp ( [tex]R_{2}[/tex] β), the equivalent resistance ( [tex]R_{eq}[/tex] β) can be calculated as follows:
[tex]\frac{1}{R_{eq}} = \frac{1}{15} + \frac{1}{8}[/tex]
Now, find the common denominator:
[tex]\frac{1}{R_{eq}} = \frac{8}{120} + \frac{15}{120} = \frac{23}{120}[/tex]
Now, take the reciprocal of both sides:
[tex]\frac{1}{R_{eq}} = \frac{120}{23}[/tex]
[tex]\frac{1}{R_{eq}}[/tex] β 5.217Ξ©
Now, to find the current (II) in the circuit, you can use Ohm's Law ( [tex]I = \frac{V}{R_{eq}}[/tex] β):
[tex]I = \frac{42}{5.217}[/tex]
I β 8.06 A
Finally, to find the current through each resistor in a parallel circuit, you can use Ohm's Law ( [tex]I_{i} = \frac{V}{R_{i}}[/tex] β):
For the bell ( [tex]I_{bell}[/tex] β):
[tex]I_{bell} = \frac{42}{15}[/tex]
[tex]I_{bell}[/tex] β 2.8 A
For the lamp ( [tex]I_{lamp}[/tex] ):
[tex]I_{lamp} = \frac{42}{8}[/tex]
[tex]I_{lamp}[/tex] β 5.25 A
