Question

A battery of 9 V is connected across the series combination of resistances 1 Ohm, 2 Ohm, and 3 Ohm. What is the value of current in 3 ohm resistor ?​

Answer 1

Answer:

1.5 A

Explanation:

We have been given:

Potential difference across the circuit =  EMF of the battery = 9V

Since all the resistances  are in series combination the current flowing though each of them will be same and only the potential difference across each will vary which means if you're able to calculate current through any/all off the resistor/s there you'll get the current though each of them

Let's begin with Net resistance in the circuit first to get current.

$\\\tt R_{net}$ =  1Ω + 2 Ω +3Ω = 6 Ω

Using Ohm's law,

V=IR

where V = potential  difference across resistor with resistance R  and has current I.

9 V = I (6 Ω)

$\\\tt I = \dfrac{9 V}{6 \Omega } \\\tt = \dfrac{3}{2} A \\\tt = 1.5 A$

This will be current though each of the resistor which includes 3 Ω.

Answer 2

Answer:

Given :-

A battery of 9 V is connected across the series combination of resistances 1 Ohm, 2 Ohm, and 3 Ohm.

To Find :-

Current

Solution :-

Since it is in series

So,

Net Resistance = R1 + R2 + R3 .... Rn

Net Resistance = 1 + 2 + 3

Net Resistance = 6 Ω

Now

I = V/R

I = 9/6

I = 1.5 A

Hence,

Current is 1.5 Ampere