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If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

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#### 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"/"R"`

= `12/6`

= 2 A

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.

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