#### 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 = *V*_{1}

From Ohm’s law, the value of *V*_{1 }can be obtained as

*V*_{1} = 2 × 1= 2 V … (i)

Potential drop across 2 Ω resistor = *V*_{2}

Again, from Ohm’s law, the value of *V*_{2 }can be obtained as

*V*_{2} = 2 × 2= 4 V … (ii)

Potential drop across 3 Ω resistor = *V*_{3}

Again, from Ohm’s law, the value of *V*_{3 }can be obtained as

*V*_{3} = 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.