Solving Problems with Kirchhoff's Laws (Cambridge (CIE) AS Physics): Revision Note

For the circuit below, state the readings of ammeters A₁, A₂  and A₃.

Answer:

Step 1: Calculate the current through ammeter A₁

• The 0.270 A and A₁ are connected in series

• The current entering the cell must equal the current out of it at A₁ 

• Therefore A₁ = 0.270 A

Step 2: Calculate the current through ammeter A₂

• From Kirchhoff's first law, the total current into a junction must equal the total current out of it

$80\mathrm{mA}\mathit{=}27\mathrm{mA}+\hspace{0.17em}{\mathrm{A}}_2$

${\mathrm{A}}_2=80-27=53\mathrm{mA}$

Step 3: Calculate the current through ammeter A₃

• Applying Kirchhoff's first law again gives:

$270\mathrm{mA}={\mathrm{A}}_3+\hspace{0.17em}80\mathrm{mA}$

$A_3=270-80=190\mathrm{mA}$

For the circuit below, state the readings of the voltmeters V₁, V₂  and V₃.

All the lamps and resistors have the same resistance.

Answer:

Step 1: Calculate the potential difference shown by voltmeter V₁

• Kirchhoff's second law states that the sum of the three components is equal to the e.m.f of the supply

$2\mathrm{V}+{\mathrm{V}}_1+12\mathrm{V}=24\mathrm{V}$

${\mathrm{V}}_1=24-12-2=10\mathrm{V}$

Step 2: Calculate the potential difference shown by voltmeter V₂

• Applying Kirchhoff's second law to this loop

• Both lamps have the same resistance R and the same current I through them so they share the energy equally and have the same potential difference

$24\mathrm{V}=2\mathrm{V}+8\mathrm{V}+\left(2{\mathrm{V}}_2\right)$

$2{\mathrm{V}}_2=24-2-8=14\mathrm{V}$

${\mathrm{V}}_2=\frac{14}{2}=7\mathrm{V}$

Step 3: Calculate the potential difference shown by voltmeter V₃

• Applying Kirchhoff's second law to this loop

$24V=2\mathrm{V}+{\mathrm{V}}_3$

${\mathrm{V}}_3=24-2=22\mathrm{V}$