Series Circuits

Q1. A simple series circuit is set up is shown using a 24 V potential difference supply from a lorry battery, an ammeter which shows a constant 1.5 A, a 14 ohm resistor and a filament bulb.

(a) What is the current passing through the filament lamp?

As you've seen previously in series circuits the same current flows through all parts of the circuit, therefore as the ammeter reads 1.5 A this is the amount of current which would be passing through the filament bulb.

(b) What is the potential difference measured by the voltmeter across the 14 ohm resistor ?

The potential difference across the resistor can be calculated using V = IR and substituting the values of I equals 1.5 and R equals 14:

In a series circuit, the potential difference supplied by the cell/battery is shared across all components. We have just calculated that the potential difference across the 14 ohm resistor would be 21 V therefore the potential difference available for the filament lamp would be 24-21 = 3 V.

(d) Finally, what is the resistance of the filament light bulb?

The resistance of the filament light can be calculated using the rearranged version of V = IR :

Q2. A simple series circuit is constructed, consisting of a single cell supplying 3 V potential difference and two resistors in series of resistance 5 ohms and 10 ohms respectively.

(a) Draw the circuit, labelled appropriately.

(b) What is the total resistance of the circuit?

In a series circuit, the total resistance is the sum of the resistances of the available components, therefore in this case the total resistance of the circuit is 15 ohms.

Given the fact that we now know the total circuit resistance of 15 ohms, and that the potential difference supplied is 3 V, we can use V = IR rearranged as I = V/R to calculate the current flowing around the circuit:

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