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Three resistors 1 ohm, 2 ohm and 3 ohm are connected in series. What is the total resistance of the combination?
If the combination is connected to a battery of emf12 V and negligible internal resistance< calculate the potential drop across each resistor.

Total resistance in series = 1+2+3= 6 ohms.
Emf (E) =12 volts and total resistance=6 ohm
therefore Current(I)= E/R =12/6 =2 amperes.
V across each resistor =
V1=I1RI=2 x 1 =2 v 
SIMILARLY V2=2x2=4 V and V3=2X3 =6 v

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Equal to 6 ohm
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Total resistance in circuit (R) = 1+2+3 =6 ohm EMF(V)= 12V Current(I)= V/R = 12/6 =2A Voltage across 1 ohm resistor = IR 1 = 2x1 = 2V ​ Voltage across 2 ohm resistor = IR2​  =​ 2x2 = 4V Voltage across 3 ohm resistor = IR 3 ​ = 2x3 = 6V​
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Total resistance in series = 1+2+3= 6 ohms.
Emf (E) =12 volts and total resistance=6 ohm
therefore Current(I)= E/R =12/6 =2 amperes.
V across each resistor =
V1=I1RI=2 x 1 =2 v 
SIMILARLY V2=2x2=4 V and V3=2X3 =6 v
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