Three wires of same material having lengths in ratio
of 2 : 3:4 and mass in ratio 3 : 4 : 5 are
connected in parallel combination across a V volt
supply, then the ratio of current flowing through
them is
(1) 54 : 64 : 75 (2) 80: 135: 192
(3) 108 : 64 : 45 (4) 75: 64 : 54
plss help me
Answers
Hi,
Answer: option (3): 108 : 64 : 45
Explanation:
The ratio of the lengths of three wires of same material is given as l₁ : l₂ : l₃ = 2 : 3 : 4 .
Also, the ratio of the masses of three wires of same material is given as m₁ : m₂ : m₃ = 3 : 4 : 5
Step 1:
We know, the resistance of a wire is calculated as,
R = ρl/A
Multiplying and dividing by “l” on R.H.S.
R = ρl²/Al = ρl²/V = [ρl²d]/m ….. [since Volume = area * length and Volume = Mass/density]
⇒ R ∝ [l²/m] ….. [since other parameters are constant] ….. (i)
Step 2:
Therefore, based on the given values and formula (i), we get
The ratio of their resistances will be,
= R₁ : R₂ : R₃
= [l₁²/m₁] : [l₂²/m₂] : [l₃²/m₃]
= [4/3] : [9/4] : [16/5] ….. (ii)
Step 3:
Since all the three wires are given to be connected in parallel, therefore the voltage across all the three resistances are same.
If the current across a resistor is given as = Voltage(V) / Resistance(R)
Thus,
The ratio of current flowing through the wires is,
= I₁ : I₂ : I₃
= [V/R₁] : [V/R₂] : [V/R₃]
= 3V/4 : 4V/9 : 5V/16 …. [substituting the values from (ii)]
Cancelling the similar terms and taking the L.C.M of the denominators
= [36*3] : [16*4] : [5*9]
= 108 : 64 : 45
Hope this is helpful!!!!