Question
If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor
Solution
Current flowing through the circuit = I
Emf of the battery, E = 12 V
Total resistance of the circuit, R = 6 Ω
The relation for current using Ohm’s law is,
`I=E/A`
`=12/6=2A`
Potential drop across 1 Ω resistor = V1
From Ohm’s law, the value of V1 can be obtained as
V1 = 2 × 1= 2 V … (i)
Potential drop across 2 Ω resistor = V2
Again, from Ohm’s law, the value of V2 can be obtained as
V2 = 2 × 2= 4 V … (ii)
Potential drop across 3 Ω resistor = V3
Again, from Ohm’s law, the value of V3 can be obtained as
V3 = 2 × 3= 6 V … (iii)
Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V respectively.
