Question

.A battery of 9V is connected in series with resistors of 0.2Ω, 0.3 Ω, 0.4 Ω &

12Ω respectively. How much current would flow through the 12Ω resistor?​

Answer 1

Answer:

For resistors in series,

R

eq

=R

1

+R

2

+R

3

+R

4

+R

5

=0.2+0.3+0.4+0.5+12

=13.4 Ω

By ohm's Law:

V=IR

eq

9=13.4I

I=0.67 A

When resistors are connected in series, current is same in all the resistors. Hence, current in 12 Ω resistor =0.67 A.

Answer 2

Given :

Potential difference, V = 9V

Resistors connected in series

• R₁ = 0.2Ω
• R₂ = 0.3Ω
• R₃ = 0.4Ω
• R$\sf{_4}$ = 12Ω

To find :

current flowing through the 12Ω resistor.

Solution :

firstly, we have to find the resistance of the series connection.

The combined resistance of any number of resistance connected in series is equal to the sum of the individual resistances. .i.e.,

➝ R = R₁ + R₂ + R₃ + R₄

➝ R = 0.2 + 0.3 + 0.4 + 12

➝ R = 12.9 Ω

now, using ohm's law .i.e.,

At constant temperature, the current through a conductor is directly proportional to the potential difference across its ends. .i.e.,

➝ V = R × I

➝ 9 = 12.9 × I

➝ I = 9/12.9

➝ I = 0.69 A

when the resistors are connected in series, the current is equal in each resistor.

thus, current flowing through the 12Ω rresistor is 0.69A.

Remember !

SI unit of Current is Ampere (A).