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Resistance in Series

When three resistors, R1, R2 and R3 are connected in series, i.e. one end to another end such that the same amount of current flow through them when they are connected to a source of e.m.f.

The total resistance, R (effective resistance) is given by V = IReff

resistance in series 2

From the circuit diagram, the voltage V is connected across the three resistors. This voltage is shared across each of the resistors.
Remember that the voltages drop across each of the resistors are different because the resistances of the resistors are different. Therefore, Let the Voltage of the cell connected across the three resistors be V.

Let the voltage drop across resistor R1 be V1.

Let the voltage drop across resistor R2 be V2, also,

Let the voltage drop across resistor R3 be V3.

Therefore the total voltage can then be found;

V = V1 + V2 + V3

IReff = \( \scriptsize \pm R1 \pm R2 \pm R3 \)

Using Ohm’s law:

V = IReff

V1 = IR1

V2 = IR2

V3 = IR3

Since the same current flows through the resistors, 

Then IReff = IR1 + IR2 + IR3

Factorising I we then have:

IReff = I(R1 + R2 + R3)

We then divide both sides by I

= \( \frac {IR_{eff}}{I} = \frac {I \left ( R_1 + R_2 + R_3 \right)}{I} \)

∴  Reff = R1 + R2 + R3

The total resistance is greater than the highest resistance in the circuit.

Example: Two resistors of value 3Ω and 5Ω are connected in series. Find the effective resistance.

resistance series

Rtotal = R1 + R2

= 3 + 5

= 8Ω


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