Question

what current will flow in a 120 volt circuit containing 3 resisitances of 1.5ohms, 3.0 ohms, and 4.5 ohms connected in parallel?

Answer 1

Hello,

Potential difference = 120 v

The resistors have resistances of
1.5 ohm, 3 ohm and 4.5 ohm

Now,
The resistors are connected in parallel.
So,
From ohms law the equivalent resistance will be equal to :-

$\frac{1}{r \: (eq)} = \frac{1}{r1} + \frac{1}{r2} + \frac{1}{r3} + \frac{1}{r4} ........$
So,

In this case,
The equivalent resistance will be equal to

$\frac{1}{r(eq)} = \frac{1}{1.5} + \frac{1}{3} + \frac{1}{4.5} \\ \\ = \frac{10}{15} + \frac{1}{3} + \frac{10}{45} \\ = \frac{2}{3} + \frac{1}{3} + \frac{2}{9} \\ \\ = \frac{6 + 3 + 2}{9} \\ = \frac{11}{9} \\ \\ r(eq) = \frac{9}{11}$

Now

The equivalent resistance is equal to 9/11 ohm

Now,
We know that

V = IR

THAT IS
Potential difference =current flowing *resistance
So,

Current flowing will be
$i \: = \frac{v}{r} \\ i = \frac{120}{ \frac{9}{11} } = \frac{120 \times 11}{9} \\ \\ i = \frac{40 \times 11}{3} = \frac{440}{3} = 146.66 \: ampere$
So,

The current flowing through the circuit is
146.66 or 146.7 (approx) Ampere.


Hope this will be helping you ✌️