Revise Kirchhoff's laws for Cambridge 9702 (10.2)

Imagine a crossroads for electrical current. Kirchhoff's First Law, often called the Current Law (KCL), tells us that the total amount of electric charge flowing into any junction in a circuit must exactly equal the total amount of charge flowing out of it. This is a direct consequence of the principle that charge is conserved – it can't accumulate or disappear at a point.

$\Sigma I_{in} = \Sigma I_{out}$

Now, consider a complete journey around any closed loop within a circuit. Kirchhoff's Second Law, the Voltage Law (KVL), states that if you add up all the potential differences (voltages) encountered along this loop, the algebraic sum must be zero. This law directly stems from the conservation of energy; no energy is gained or lost when you return to your starting point. It can also be stated as: the sum of the e.m.f.s in a closed loop equals the sum of the potential drops.

$\Sigma V = 0 \quad \text{or} \quad \Sigma \epsilon = \Sigma IR$

Correctly applying KVL depends on a consistent sign convention. This is a common source of errors, so follow these steps carefully:

To apply Kirchhoff's Second Law effectively, you need to be familiar with a few key concepts. The electromotive force (e.m.f., \epsilon) is the total energy supplied by a source per unit charge. However, power sources aren't perfect; they have internal resistance ($r$) which dissipates some energy internally.

$\epsilon = \frac{W}{Q}$

When current flows, a potential drop occurs across this internal resistance, known as lost volts ($v = Ir$). The actual voltage delivered to the external circuit is the terminal potential difference ($V$). The relationship is straightforward: the total e.m.f. is the sum of the terminal potential difference and the lost volts.

$\epsilon = V + v \quad \Rightarrow \quad V = \epsilon - Ir$ $\text{Since } V = IR_{external} \text{ and } v = Ir, \text{ we get } \epsilon = I(R_{external} + r)$