Question

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?​

Answer 1

Explanation:

In series connection, there is no division of current. The current flowing across all the resistors is the same.

To calculate the amount of current flowing across the resistors, we use Ohm’s law.

But first, let us find out the equivalent resistance as follows:

↪R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω

↪ R = 13.4 Ω

Now, using Ohm’s law,

I = V/R

↪ I = 9/13.4

↪ I = 0.671 A

The current flowing across the 12 Ω is 0.671 A.

Answer 2

Question :

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Given: Voltage of Battery = 9V

Resistors connected in series = 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω

To find: Current flowing through 12 Ω resistor.

Solution: The sum of the resistances will give the value of R.

R= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ohm

Using,Ohms law :V= IR

I= V/R

putting the values, I= 9/13.4 = 0.671 A

Therefore, the current that would flow through the 12 Ohm resistor is 0.671 A.

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What is Series combination ?

Two or more resistors are said to be in connected in series combination, if same current passes through each of them, when some potential difference is applied across the combination.

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