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Question

Calculate the effective resistance in the following circuit.

A
5 Ω
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B
12 Ω
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C
8 Ω
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D
25 Ω
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Solution

The correct option is D 25 ΩGiven: VoltageV = 6 V Resistances, R1=5 Ω, R2=8 Ω, R3=12 Ω Since resistances are connected in series, current I through each resistance is same. Voltage gets divided among each resistance in series connection. Let Reff be the effective resistance of the circuit ∴IReff=IR1+IR2+IR3 (ohm's law) ⇒Reff=R1+R2+R3 ⇒Reff=5 Ω+8 Ω+12 Ω=25 Ω Hence, effective resistance of the ciruit=25 Ω

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