A Battery of 9V is connected in series with resistors 0.2 , 0.3, 0.4 m, 0.5 , 10 . How much current would flow through 10 resistor?
Answers
Answer:
A battery of 9 V is connected in series with resistors of 0.2 Ohms, 0.3 Ohms, 0.4 Ohms, 0.5Ohms and 12Ohms respectively
Find out
Amount of the current flow through the 12 Ω resistors
Solution
According to Ohm’s law
V= IR
Therefore, I= V/R
Where,
R is the equivalent resistance of resistances 0.2 Ohm, 0.3 Ohm, 0.4Ohm, 0.5 Ohm and 12 Ohm.
These are connected in series.
Hence, the sum of the resistances will give the value of R.
R= R1 + R2 + R3 + R4 + R5
= 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ohm
The potential difference, V= 9 V
I = 9/13.4 = 0.671 A
When resistors are connected in series, the current is the same in all the resistors. Hence, current in 12 Ω resistor = 0.67 A.
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Answer:
0.78 A current flows through 10 Ω.
Explanation:
Equivalent Resistance (R) = R₁ + R₂ + R₃ + R₄ + R₅ = 0.2 + 0.3 + 0.4 + 0.5 + 10
= 11.4 Ω
Potential Difference (V) = 9 V
Current (I) = Potential difference (V) / Equivalent Resistance (R)
= 9 / 11.4
= 0.78 A
When resistors are connected in series, current flows through each resistor is same.
∴ 0.78 A current flows through 10 Ω resistor.