Question

three resistors of 3ohm each are connected to a battery of 3volt where two are connected in series andthird one is parallel to them. calculate the current drawn from battery.

Answer 1

Hello,

The resistance of the resistors are
3 OHM each.

Two are connected in series.
Say r1 and r2 are connected in series connections.

So, from OHM'S Law,
The equivalent resistance of r1 and r2 is

$r(eq) = r1 + r2 + r3 + r4 +.....rn$
So,
R1 and r2 have resistance of 3 ohm each
So
Equivalent resistance is equal to

$r(eq) = 3 + 3 = 6 \: \: ohm$
Now,

Another resistor say r3 is connected parallel in the series.

So,

Equivalent resistance of the r3 and r(eq) of r1 and r2

From ohms law we know that

$\frac{1}{r(eq)} = \frac{1}{r1} + \frac{1}{r2} + .......$
In this case puffing values.

$\frac{1}{r(eq)} = \frac{1}{6} + \frac{1}{ 3 } \\ \\ = \frac{3}{6} = \frac{1}{2} \\ \\ r(eq) = 2 \: \: ohms$

Now,

We know that
$i = vr$
Where i is the current drawn from the circuit and v is the potential difference and r is the resistance offered.
And in this case r is the final equivalent resistances of the 3 resistors.

So

Putting the values
$i = vr \\ \\ i = 3 \times 2 = 6 \: \: ampere$

Hope this will be helping you.....