Kirchhoff’s First Law
Kirchhoff’s first law is a consequence of the conservation of charge and states that:
The sum of currents entering a node (or junction) is equal to the sum of the currents leaving a node.
A node is a point where two or more circuit elements are connected. Let’s look at the circuit below:
Here we can see that the current entering the first node has a value of 10A. The circuit then splits into two wires and so the current also divides into two. Some of the electrons flow into wire x and some into wire y.
The diagram shows that the value of the current in wire x is 8A. From Kirchhoff’s first law we know that the value of the current entering the node must be equal to the the sum of the currents leaving the node. Therefore, if we have 10A entering the node, and one of the wires has a current of 8A the current in wire y must be 2A (10 – 8 = 2).
The current can be split into many directions (not just two) and the same law applies.
Kirchhoff’s Second Law
Kirchhoff’s second law is also a consequence of the conservation of charge but also a consequence of the conservation of energy and states:
The sum of all the potential differences (voltages) in a closed loop in a circuit is equal to zero.
As charge carries (such as electrons) pass through a component work is done on them and they either gain or lose electrical energy. Charge carriers gain electrical energy from the source (such as a battery or cell) and lose energy when they pass through a component (such as a resistor or bulb).
If we state that a rise in voltage is positive (the charge carriers passing through a cell) and a drop in voltage as negative (the charge carriers passing through a component) the sum will be zero if the loop is closed.
A simpler way of stating Kirchhoff’s second law is:
The sum of the electromotive forces in a closed loop in a circuit must be equal to the sum of the potential differences (voltages) in the closed loop.
Let’s look at the circuit below:
The e.m.f. (ε) of the circuit is 20V and we can see that the potential difference across resistor x is 5V. From Kirchhoff’s second law we know that the e.m.f. must be equal to the sum of the potential differences. Therefore, if the e.m.f. is 20V and the p.d. across one of the resistors is 5V the potential difference across resistor y must be 15V (20 – 5 = 15).
Both the first and second law are useful when analysing series and parallel circuits.
Worked Examples
Example 1
In the circuit below, state the current flowing through resistor B and resistor C.
Example 2
In the circuit below, state the value of the potential difference across resistor B.
Example 3
In the circuit below, state the value of the potential difference across resistor B.
This question requires knowledge of current, potential difference and resistance.
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