a battery of 9v is connected in series with resistor of 0.2ohm, 0.3ohm,0.5ohm and 11ohm respectively. How much current would flow through 11 ohm resistor?
Answers
GiveN:-
A battery of 9 V is connected in series with resistor of 0.2 Ω, 0.3 Ω, 0.5 Ω and 11 Ω respectively.
To FinD:-
How much current would flow through 11 Ω resistor?
SolutioN:-
Analysis :
Here the concept of combination of series resistance is used. First we have to find the equivalent resistance for the 4 resistors and then by using Ohm's Law we can find the current.
Solution :
We know that in series combination, the resistors are joined one after the other. So for equivalent resistance we have to add up all the resistors together.
We know that if we are given the respective resistors and is asked to find the equivalent resistance for the series combination then our required formula is,
R = R₁ + R₂ + R₃ + R₄ + ...... so on
where,
- R is the equivalent resistance
- R₁ = 0.2 Ω
- R₂ = 0.3 Ω
- R₃ = 0.5 Ω
- R₄ = 11 Ω
Using the required formula and substituting the required values,
⇒ R = R₁ + R₂ + R₃ + R₄
⇒ R = 0.2 Ω + 0.3 Ω + 0.5 Ω + 11 Ω
⇒ R = 12 Ω
∴ Resistance = 12 Ω.
Now the current :
By using Ohm's Law,
V = IR
where,
- V is Potential Difference
- I is Current
- R is Resistance
We know that if we are given the Potential Difference and Resistance and asked to find the current then our required formula is,
I = V/R
where,
- V = 9 V
- R = 12 Ω
Using the required formula and substituting the required values,
⇒ I = V/R
⇒ I = 9/12
⇒ I = 3/4
⇒ I = 0.75 A
∴ I or Current = 0.75 A.
Current that would flow through 11 Ω resistor is 0.75 ampere(A).
Explore More :
- Ohm's Law:-
Ans)) According to Ohm's Law, the current flowing in a conductor is directly proportional to the potential difference applied across its ends provided that the physical conditions and the temperature of the conductor remain constant.
I ∝ V
⇒ V = IR